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One cone, four curves

◆FRAME

Point a flashlight straight at a wall. The beam lands as a circle. Tilt the flashlight back a little, and the circle stretches into an oval. Tilt it further still, and the oval opens up into a curve that never closes.

Three different shapes came from one cone of light hitting one wall. Only the angle changed.

A CONIC SECTION makes this idea mathematical. Slice a cone with a flat plane, and the cut edge traces one of four curves: a CIRCLE, an ELLIPSE, a PARABOLA, or a HYPERBOLA. Which curve you get depends only on the angle of the slice.

The four curves look unrelated as four separate equations. Underneath, they come from one shape, cut differently. The differences between their equations are not arbitrary. They record exactly how the slicing angle changed.

The fourth curve needs one thing a single flashlight beam cannot give: a cone has two mirror-image halves, meeting at the same point. Only the hyperbola needs both halves cut by the same plane.

One cone, sliced at different angles, produces every curve this chapter studies.

You met all four of these long before you met their equations, and it is worth saying so before the fourteen sections after this one hand you four different-looking formulas. Two details here are paid off later. The sun is drawn off the centre of its orbit — the hollow dot is the centre, the orange one is the sun — and that gap is the whole of what makes an ellipse an ellipse rather than a circle. And the cooling tower is the only one of the four that needs two separate pieces of curve, which is exactly why the cone on the next page has a second half.
This is where the family gets its name: circle, ellipse, parabola and hyperbola are literally slices of one cone, cut at four different angles. The cone itself never appears again in this chapter — every curve ahead is instead defined algebraically, as a locus of points obeying a distance rule from a focus, and both definitions, geometric and algebraic, land on the exact same four shapes.

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Eccentricity: the number that names the curve

●KEY-TERM

Four curves, four different definitions on the page. Is there one number that sorts them all?

There is. It is called ECCENTRICITY, written $e$. One number tells you exactly which curve an equation describes.

$e = 0$ gives a CIRCLE. $0 < e < 1$ gives an ELLIPSE. $e = 1$ gives a PARABOLA. $e > 1$ gives a HYPERBOLA.

The four curves are not four separate rulebooks. They are one family, ordered along a single number line.

Watch the number move. At $e=0$ the curve is a perfect circle. Nudge it to $e=0.3$: a mildly flattened ellipse. Push it to $e=0.9$: a long, narrow ellipse, almost ready to open up entirely. Cross exactly $e=1$, and the curve stops closing on itself — it becomes a parabola. Push further still, to $e=1.5$ or higher, and the curve splits into the hyperbola’s two separate branches. Nothing about the underlying definition changed; only one number did.

Eccentricity is the single number that tells you which of the four curves you are looking at.

Four curves, four separate definitions — and one number that puts them in order. Walk from 0 upward. At 0 the two foci have met, so c = 0 and the curve closes into a circle. Anywhere between 0 and 1 it is an ellipse, and the further you walk the thinner it gets. Exactly at 1 it stops closing and runs off as a parabola. Past 1 it has broken into a hyperbola’s two branches, and there is no upper limit to walk to. The two worked values are the chapter’s own curves, and they are the same 3-4-5 triangle: put a on the hypotenuse and e = 3/5 lands below 1; put c there and e = 5/3 lands above it.

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The circle: one fixed distance

▲CONCEPT

Start with the simplest curve on the eccentricity line: $e = 0$, the CIRCLE.

A circle is the set of every point at a fixed distance from one fixed point. The fixed point is the CENTRE. The fixed distance is the RADIUS, $r$.

Test the definition on numbers. Centre the circle at the origin with radius $5$. Does $(3,4)$ lie on it? Its distance from the origin is $\sqrt{3^2+4^2} = \sqrt{25} = 5$ — exactly the radius, so the point sits right on the circle, neither inside it nor outside.

Put the centre at $(h, k)$. A point $(x, y)$ lies on the circle exactly when its distance to $(h,k)$ equals $r$. Squaring the distance formula to clear the square root gives the STANDARD EQUATION of a circle:

$(x-h)^2 + (y-k)^2 = r^2.$

Every circle’s equation has exactly this shape. Only $h$, $k$, and $r$ change from one circle to the next.

A circle is one distance, held fixed, in every direction at once.

This triangle is what the general form ahead trades away: expand (x − h)² + (y − k)² = r² and h, k, r scatter into three coefficients that no longer look like a centre or a radius. Reading a circle from its standard form is just reading this triangle off the page; reading it from the general form means rebuilding the triangle first.
Worked example

Find the equation of the circle with centre $(2, -3)$ and radius $5$

  1. centre $(h,k) = (2,-3)$, radius $r = 5$
    the two numbers the standard circle equation needs
  2. $(x-2)^2 + (y-(-3))^2 = 5^2$
    substitute directly into $(x-h)^2+(y-k)^2=r^2$
  3. $(x-2)^2 + (y+3)^2 = 25$
    simplify the double negative and square the radius
The worked example above is handed the two numbers it needs. An exercise usually is not: it gives a point the circle passes through, or the two ends of a diameter, and the first move is to get back to a centre and a radius. Panel 2 does it with the distance formula, panel 3 with a midpoint and a half-length. After that the three problems are the same problem, and the last step is one substitution.

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The circle’s general equation

▲CONCEPT

Expand the standard equation’s brackets, and a different-looking form appears.

$(x-h)^2 + (y-k)^2 = r^2$ expands to $x^2 - 2 h x + h^2 + y^2 - 2 k y + k^2 = r^2$. Collect terms, and write $-2h$ as $2 g$ and $-2k$ as $2 f$. What remains is the GENERAL EQUATION of a circle:

$x^2 + y^2 + 2 g x + 2 f y + c = 0.$

Every circle can be written this way. The centre and radius are still there — just hidden inside $g$, $f$, and $c$ instead of $h$, $k$, and $r$.

Undo the substitution, and the centre and radius come straight back out: centre $= (-g, -f)$, radius $= \sqrt{g^2 + f^2 - c}$. No completing the square by hand is needed once the equation is already written this way — the coefficients hand the answer over directly.

Same circle, different-looking equation. The centre and radius are still sitting in the coefficients, one algebra step away.

This general form matters less for reading a circle already given than for building one from scratch: hand the algebra three conditions instead — three points it must pass through, say, or a point plus a tangent line — and g, f and c fall out as the unknowns of three linear equations. That three-condition route, not the centre-radius form, is how most general-equation circle problems actually start.
Worked example

Find the centre and radius of $x^2 + y^2 - 4 x + 6 y - 12 = 0$

  1. $2 g = -4$, so $g = -2$
    match the x-coefficient to $2 g$ in the general equation
  2. $2 f = 6$, so $f = 3$; and $c = -12$
    match the y-coefficient to $2 f$ the same way, and read off the constant term
  3. centre $= (-g, -f) = (2, -3)$
    the general equation’s own centre formula, read straight off $g$ and $f$
  4. radius $= \sqrt{g^2+f^2-c} = \sqrt{4+9+12} = \sqrt{25} = 5$
    substitute $g$, $f$, and $c$ into the radius formula
The algebra above ends at two numbers. This is where you find out whether they are the right two. Neither test below uses g, f or c again — one walks a 3-4-5 triangle out from the claimed centre and lands on the curve, the other puts y = 0 back into the equation as it was first written and asks where the curve should cross. Make a habit of the second one: it costs a factorised quadratic and it catches the sign slip that the readoff is most likely to make.

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The parabola: a point and a line, balanced

▲CONCEPT

Move up the eccentricity line to $e = 1$: the PARABOLA.

A parabola is the set of every point equally distant from one fixed point and one fixed line. The fixed point is the FOCUS. The fixed line is the DIRECTRIX, and it never passes through the focus.

This shape carries one sharp geometric property, provable straight from the definition: a ray travelling parallel to the axis, after bouncing off the curve, always passes through the focus, no matter where along the curve it strikes. Turn that same idea around, and every ray leaving the focus emerges travelling parallel to the axis — the distance-equality in the definition is exactly what forces this, since the incoming and outgoing angles are both measured against the same fixed distance to the directrix.

Two more terms name the parabola’s own landmarks. The AXIS is the line through the focus, perpendicular to the directrix. The VERTEX is the point where the parabola crosses its own axis — exactly midway between the focus and the directrix.

A parabola holds one distance equal to another: point to focus, point to line.

Every curve after the circle in this chapter repeats this same move: pick which distances have to stay equal, then let the algebra follow from that. The parabola is the plain version, one point and one line; the ellipse and hyperbola ahead swap the line for a second point, and equal for constant sum or constant difference.

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The parabola’s standard equation

▲CONCEPT

Put the vertex at the origin and the axis along the positive x-axis. The focus sits at $(a, 0)$ for some $a > 0$, and the directrix is the vertical line $x = -a$.

Every point on the parabola sits the same distance from $(a,0)$ as it does from the line $x=-a$. Writing that distance equality out and simplifying gives the STANDARD EQUATION:

$y^2 = 4 a x \cdot$

One more measurement comes free once $a$ is known. The LATUS RECTUM is the focal chord perpendicular to the axis — the widest gap between the two branches, measured straight through the focus. Its length is always $4 a$, the same $4 a$ already sitting on the equation’s right-hand side.

A parabola can open left, up, or down instead of right. Each of those three cases is the same equation, with a sign flipped or $x$ and $y$ swapped. This chapter works through the right-opening case in full — the pattern the other three repeat.

_Once $a$ is known, the focus, the directrix, and the latus rectum all fall out of one equation._

The single number a is doing three jobs at once in this figure: it is the vertex-to-focus distance, the vertex-to-directrix distance, and, through 4a, the width of the curve straight through the focus. Handed only that width — the way y² = 12x hands you 4a = 12 — you can still recover all three.
Worked example

Find the focus, directrix, and latus rectum of $y^2 = 12 x$

  1. $4 a = 12$, so $a = 3$
    match the given equation to the standard form $y^2 = 4 a x$
  2. focus $= (a, 0) = (3, 0)$
    the standard equation’s own focus formula, once $a$ is known
  3. directrix: $x = -a = -3$
    the directrix always sits the same distance $a$ on the opposite side of the vertex
  4. latus rectum $= 4 a = 12$
    the same $4 a$ that already appeared on the equation’s own right-hand side
Watch this formula change further into the chapter: the ellipse and hyperbola’s latus rectum is 2b²/a, not 4a, because each of those curves has a second axis length to draw on. A parabola has none, so its chord answer collapses to one letter, a, doing every job at once.

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The ellipse: two foci, a constant sum

▲CONCEPT

Continue up the eccentricity line, into $0 < e < 1$: the ELLIPSE.

An ellipse is the set of every point whose distances to two fixed points add up to one constant total. The two fixed points are the FOCI. The constant total is written $2 a$.

Test the definition on numbers. Put the two foci at $(-3, 0)$ and $(3, 0)$, and fix the constant total at $2a = 10$. The point $(0, 4)$: distance to each focus is $\sqrt{3^2+4^2} = 5$, so the two distances sum to $10$. The point $(5, 0)$: distances are $2$ and $8$, again summing to $10$. Two very different points, trading distance to one focus for distance to the other, landing on the identical total both times.

Every point on the curve trades distance to one focus for distance to the other. Move closer to one focus, and the point moves further from the second one, by exactly the same amount. Only the total stays fixed.

An ellipse is two distances, always adding to the same number.

The parabola one section back fixed one distance against one line; here two distances trade against each other, and it is their total, not either one alone, that stays fixed at 2a. That change, from one rule to two, is what turns a curve that runs off straight into one that closes on itself.

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The ellipse’s standard equation

▲CONCEPT

Centre the ellipse at the origin, with its major axis running along the x-axis. Call the SEMI-MAJOR axis $a$ and the SEMI-MINOR axis $b$, with $a > b > 0$. The gardener’s string-and-pegs curve, written as an equation, is:

$x^2/a^2 + y^2/b^2 = 1.$

The two foci — the two pegs — sit at $(\pm c, 0)$, where $c^2 = a^2 - b^2$.

That relationship is not a coincidence. The pegs must sit strictly inside the loop of string, or the string could never form a closed curve around them. So $c$ is always smaller than $a$, and $a^2 - b^2$ always comes out positive.

_The major axis sets the scale; $b$ and $c$ split the rest of it between the curve’s width and the foci’s spacing._

Everything later in this chapter divides by a or square-roots a squared minus b squared, and both moves quietly assume the answer stays positive. It does, and the right-hand panel is the reason. Push the two pegs apart without lengthening the string and the oval gets narrower; push them all the way to the ends and there is nothing left to be narrow. The string is still exactly long enough to reach from one peg to the other and back, so every point that satisfies the definition has been squeezed onto the straight line between them. A peg further out than that would need the string to be shorter than the gap it has to span, which no point can do. So the foci are not merely usually inside the curve — they cannot be anywhere else, and that is what keeps the subtraction under the square root from ever going negative.
Every point on the ellipse obeys the same sum-of-distances rule from the definition, but only at the top of the minor axis do the two distances become equal to each other. That one special point is what turns the general definition into this clean right triangle relating a, b and c — nowhere else on the curve does the triangle appear this simply.

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Ellipse eccentricity and the latus rectum

▲CONCEPT

The ellipse’s own eccentricity is $e = c/a$.

Since $c$ is always smaller than $a$, this ratio always lands strictly between 0 and 1 — exactly the ellipse’s own slot on the eccentricity line from earlier in this chapter.

A more stretched ellipse has $c$ closer to $a$, so $e$ climbs closer to 1. A rounder ellipse has $c$ closer to 0, so $e$ sits closer to 0 as well.

One more measurement completes the picture: the LATUS RECTUM, the focal chord perpendicular to the major axis. Its length is $(2 b^2)/a$.

A more stretched ellipse has eccentricity closer to 1; a rounder one has eccentricity closer to 0.

Notice that no new equation was solved to reach either number here — both came straight off a, b and c, already known from the standard form. That is the pattern worth banking: once a, b and c are pinned down, everything else this section names is a one-line readout, not a fresh derivation. An exam asking for ‘the eccentricity and the latus rectum’ is asking you to read, not solve.
Worked example

Find the foci, eccentricity, and latus rectum of $x^2/25 + y^2/16 = 1$

  1. $a^2 = 25$, $b^2 = 16$, so $a = 5$, $b = 4$
    read the two denominators straight off the standard equation
  2. $c = \sqrt{a^2-b^2} = \sqrt{25-16} = 3$
    the ellipse’s own relation between $a$, $b$, and $c$
  3. foci $= (\pm 3, 0)$
    the foci sit at $(\pm c, 0)$ once $c$ is known
  4. eccentricity $e = c/a = 3/5$
    the ratio that always lands between 0 and 1 for an ellipse
  5. latus rectum $= (2 b^2)/a = 32/5$
    substitute $a$ and $b$ into the ellipse’s latus rectum formula

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The hyperbola: two foci, a constant difference

▲CONCEPT

Move to the far end of the eccentricity line, past $e = 1$, into $e > 1$: the HYPERBOLA.

A hyperbola is the set of every point where the difference of the distances to two fixed points stays constant — but only once that difference is stripped of its sign. The two fixed points are again called the FOCI. The constant is again written $2 a$.

The absolute value matters here in a new way. A hyperbola has two separate branches. On one branch, the point sits closer to one focus. On the other branch, it sits closer to the other focus. Which distance is larger keeps swapping from branch to branch. Only the unsigned difference stays constant across the whole curve.

Test the branch-swap behaviour directly, using plain distances rather than coordinates. Suppose one point sits $2$ units from the first focus and $8$ units from the second focus — a difference of $6$. A point on the OTHER branch might sit $9$ units from the first focus and $3$ units from the second — again a difference of $6$, only now it is the first distance that is larger. The unsigned difference stays fixed at $6$ on both branches; which raw distance is bigger flips.

A hyperbola holds a difference fixed, not a sum — and that one sign change is what splits the curve into two branches.

The page says a hyperbola has two branches. What it does not say is what decides where a branch begins, or why the middle of the picture is empty, and both fall out of one walk. Start at the centre, where the two distances are equal and their difference is nothing at all, and move right. The difference grows the whole way: the near focus keeps getting nearer and the far one keeps getting further, so every step counts twice. It passes 6 exactly once, and the point where it does is what we have been calling the vertex. Everything the walk crossed on the way there had a difference too small to qualify, which is the gap between the branches — not a gap in the drawing, a range of values the definition refuses. Everything past it has a difference too large. Notice also where the climb stops: at 10, the distance between the foci themselves, since no point can beat a straight line. That ceiling is why the constant has to be smaller than the gap between the foci before there is any curve to draw.
This is the ellipse’s rule with the sign changed: the same two foci, the same idea of a constant, but now it is how much farther one focus sits than the other, not how far both sit combined. Sum to difference is the entire distinction between the two curves’ shapes — everything else in this chapter follows from that one change.

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The hyperbola’s standard equation

▲CONCEPT

Centre the hyperbola at the origin too, with its TRANSVERSE axis along the x-axis. The equation looks almost identical to the ellipse’s — except for one sign:

$x^2/a^2 - y^2/b^2 = 1.$

The foci again sit at $(\pm c, 0)$, but now $c^2 = a^2 + b^2$ — a plus, where the ellipse had a minus.

That single sign change is the whole story. The ellipse’s minus sign forced $c$ to stay smaller than $a$, since $b^2$ had to be subtracted away. The hyperbola’s plus sign removes that ceiling entirely: $c$ is always GREATER than $a$ here, however large $b$ grows.

One sign, flipped, is the entire algebraic difference between a curve that closes on itself and one that never does.

A focus sitting outside the curve is why a hyperbola’s two branches open away from each other instead of closing around a shared region the way an ellipse’s foci, sitting inside, let it close around both. Get this backwards while sketching — a focus drawn inside a branch — and the curve you have drawn cannot actually be a hyperbola.

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Hyperbola eccentricity and the latus rectum

▲CONCEPT

The hyperbola’s own eccentricity is the same ratio as before: $e = c/a$.

Since $c$ is always greater than $a$ here, this ratio always lands above 1 — the mirror image of the ellipse’s own $e < 1$. The two curves sit on opposite sides of $e = 1$, using the identical formula for $e$.

The LATUS RECTUM formula carries over too: length $= (2 b^2)/a$, the same shape as the ellipse’s, built now from the hyperbola’s own $a$ and $b$.

_The ellipse and the hyperbola share the exact same eccentricity formula; only which side of $e=1$ they land on differs._

Unlike the ellipse, e has no ceiling here: a hyperbola can sit barely open, just above e = 1, or run out nearly straight, at e far larger, and nothing in the definition caps it the way e < 1 caps the ellipse. That missing ceiling is the algebraic fingerprint of a sign flip — b² = c² − a² here, against b² = a² − c² there — which is what lets c run arbitrarily far past a in the first place.
Worked example

Find the foci, eccentricity, and latus rectum of $x^2/9 - y^2/16 = 1$

  1. $a^2 = 9$, $b^2 = 16$, so $a = 3$, $b = 4$
    read the two denominators straight off the standard equation
  2. $c = \sqrt{a^2+b^2} = \sqrt{9+16} = 5$
    the hyperbola’s own relation between $a$, $b$, and $c$ — a plus, not the ellipse’s minus
  3. foci $= (\pm 5, 0)$
    the foci sit at $(\pm c, 0)$ once $c$ is known
  4. eccentricity $e = c/a = 5/3$
    the ratio that always lands above 1 for a hyperbola
  5. latus rectum $= (2 b^2)/a = 32/3$
    the same formula shape as the ellipse’s, built from the hyperbola’s own $a$ and $b$

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Every curve, one question

■RECAP

Every curve in this chapter answers one question: how does a point’s distance behave?

A circle holds ONE distance fixed — to the centre. An ellipse holds the SUM of two distances fixed — to the two foci. A hyperbola holds the ABSOLUTE DIFFERENCE of two distances fixed — to the two foci again. A parabola holds distance to one point equal to distance to one line.

Eccentricity is the single number that names which behaviour a given equation encodes: $e=0$, or $0<e<1$, or $e=1$, or $e>1$. Once the curve’s type is known, every constant reads off the equation the same direct way, every time. That constant is $r$ for a circle, $a$ for a parabola, and $a$ and $b$ for an ellipse or a hyperbola.

In practice, start from whichever fact a problem hands you. Told a fixed distance from one point? That is a circle. A constant sum of distances to two points? An ellipse. A constant, unsigned difference of distances? A hyperbola. Distance to a point equal to distance to a line? A parabola. Eccentricity then confirms which regime an equation already sits in, once it is written down.

One cone, four ways of holding distance fixed, one number that tells you which.

Each of these four constructions has been drawn once already, alone, pages apart. The recap’s claim is the thing none of those four could show on its own: they are one question with four answers. The ellipse and the hyperbola here are drawn deliberately with the same pair of foci — nothing moved between those two panels except which arithmetic is done with the two bars, a sum or a difference, and that alone is what sends e below 1 or above it.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. The relationship between $a$, $b$, and $c$ uses the same formula for an ellipse and a hyperbola. Just remember that $c^2$ equals $a^2$ plus or minus $b^2$, and guess the sign.

THE REALITY. The sign is not a coin flip. It comes straight from each curve’s own definition. An ellipse’s two foci sit INSIDE the curve, which forces $c^2 = a^2 - b^2$, so $c < a$. A hyperbola’s two foci sit OUTSIDE the curve, which forces $c^2 = a^2 + b^2$, so $c > a$.

Use the ellipse’s minus sign on a hyperbola, or the hyperbola’s plus sign on an ellipse, and the $c$ that comes out contradicts the very curve it claims to describe.

The sign is not a detail to memorise separately — it is the algebra reporting where the foci actually sit.

The exam version of this mix-up is memorising ‘ellipse minus, hyperbola plus’ as two separate facts and mixing them up under time pressure. They were never two facts: redraw the triangle, and whichever letter is the longest side is the one that squares alone on its own side of the relation — the sign follows the drawing, not the other way round.
Take the guess seriously for a moment and draw what it asks for. Give the ellipse the hyperbola’s plus sign and c comes out at 6.40, which is the hollow ring — past the vertex, off the curve, in the one place an ellipse’s focus can never be. Give the hyperbola the ellipse’s minus sign and it asks for the square root of minus 7, so there is nothing to mark at all. The right answer is not the one you remember; it is the one that leaves the focus where the curve can hold it.
✕MISCONCEPTION

THE TRAP. In $x^2/a^2 + y^2/b^2 = 1$, the LARGER denominator is always the one sitting under $x^2$.

THE REALITY. $a$ is defined as the SEMI-MAJOR axis — the larger one — by convention. That part is fixed. But which position, under $x^2$ or under $y^2$, actually holds the larger number depends on which axis the ellipse’s major axis runs along.

This chapter’s own scope fixes the major axis along the x-axis throughout. So inside this chapter, $a^2$ really is always the denominator under $x^2$ — the habit is not wrong here. It breaks the moment a problem places the major axis along the y-axis instead, which swaps which denominator is larger without changing what the letter $a$ itself means.

_$a$ always means the larger axis. Which position it sits under is a fact about the ellipse, not a fact about the letter._

This particular trap is specific to the ellipse. The hyperbola earlier in this chapter carries no version of it: its equation already has a minus sign fixed to one term, so there is nothing to decide. Here both denominators are positive, and only their size — never which variable sits underneath — settles which one is really a².

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Practice set

Exercise 10.1 — The circle
  1. practice Find the equation of the circle with centre $(3, -2)$ and radius $6$.
  2. practice Find the equation of the circle centred at the origin and passing through $(5, 12)$.
  3. practice Find the centre and radius of the circle $x^2 + y^2 - 6 x + 8 y - 11 = 0$.
  4. practice Which of these equations represents a circle centred at the origin?
    1. $x^2 + y^2 = 25$
    2. $(x-2)^2 + y^2 = 25$
    3. $x^2 + (y-3)^2 = 25$
    4. $(x-1)^2 + (y-1)^2 = 25$
  5. practice A circle has a diameter with endpoints $(2, 3)$ and $(-2, 5)$. Find its equation.
Answers
  1. $(x-3)^2 + (y+2)^2 = 36$
  2. $x^2 + y^2 = 169$
  3. centre $= (3, -4)$, radius $= 6$
  4. $x^2 + y^2 = 25$
  5. $x^2 + (y-4)^2 = 5$
Exercise 10.2 — The parabola
  1. practice Find the focus, directrix, and latus rectum of $y^2 = 8 x$.
  2. practice Find the equation of the parabola with vertex at the origin, opening along the positive x-axis, and passing through $(4, 8)$.
  3. practice A parabola has equation $y^2 = 20 x$. What is the length of its latus rectum?
    1. $5$
    2. $10$
    3. $20$
    4. $40$
  4. practice Find the equation of the parabola with focus $(6, 0)$ and directrix $x = -6$.
Answers
  1. focus $= (2, 0)$, directrix $x = -2$, latus rectum $= 8$
  2. $y^2 = 16 x$
  3. $20$
  4. $y^2 = 24 x$
This exercise asks for parabolas the chapter never drew. Only one orientation appears in the text, y² = 4ax, and the rest were left to prose. The two things worth reading off the row: the letter that is SQUARED names the axis the curve opens along, and the sign in front of 4a picks which end of that axis. Everything else — where the focus sits, how far the directrix stands on the other side, how wide the curve is at the focus — is a = 1 and 4a = 4 in every panel.
Exercise 10.3 — The ellipse
  1. practice Find the foci, eccentricity, and latus rectum of $x^2/36 + y^2/20 = 1$.
  2. practice Find the equation of the ellipse with foci $(\pm 5, 0)$ and vertices $(\pm 13, 0)$.
  3. practice Find the eccentricity of the ellipse $4 x^2 + 9 y^2 = 36$.
  4. practice An ellipse has eccentricity $e = 4/5$. Which is true?
    1. $e$ is impossible for an ellipse
    2. $0 < e < 1$, so this is a valid ellipse
    3. $e = 1$ means this is a parabola
    4. $e > 1$ means this is a hyperbola
  5. practice Find the lengths of the major and minor axes of the ellipse $x^2/49 + y^2/9 = 1$.
  6. practice Find the equation of the ellipse with semi-major axis $10$, semi-minor axis $6$, and major axis along the x-axis.
  7. practice Find the length of the latus rectum of the ellipse $x^2/100 + y^2/64 = 1$.
Answers
  1. foci $= (\pm 4, 0)$, $e = 2/3$, latus rectum $= 20/3$
  2. $x^2/169 + y^2/144 = 1$
  3. $e = \sqrt{5}/3$
  4. $0 < e < 1$, so this is a valid ellipse
  5. major axis $= 14$, minor axis $= 6$
  6. $x^2/100 + y^2/36 = 1$
  7. $64/5$
The exercise asks for an eccentricity in one question, a latus rectum in another and the axis lengths in a third, as if they were three separate things to remember. They are one dial. Fix the major axis and turn e up: the focus walks outward, the curve flattens, and the chord standing on the focus gets shorter with it. The left panel is the setting the chapter names but never draws — at e = 0 the two foci have met at the centre and the "chord on the focus" is just the diameter.
Exercise 10.4 — The hyperbola
  1. practice Find the foci, eccentricity, and latus rectum of $x^2/16 - y^2/9 = 1$.
  2. practice Find the equation of the hyperbola with foci $(\pm 13, 0)$ and vertices $(\pm 5, 0)$.
  3. practice A hyperbola has eccentricity $e = 3/2$. Which is true?
    1. $0 < e < 1$ (an ellipse)
    2. $e = 1$ (a parabola)
    3. $e > 1$, so this is a valid hyperbola
    4. impossible for any conic
  4. practice Find the eccentricity of the hyperbola $16 x^2 - 9 y^2 = 144$.
  5. practice Find the eccentricity and latus rectum of the hyperbola $x^2/4 - y^2/12 = 1$.
Answers
  1. foci $= (\pm 5, 0)$, $e = 5/4$, latus rectum $= 9/2$
  2. $x^2/25 - y^2/144 = 1$
  3. $e > 1$, so this is a valid hyperbola
  4. $e = 5/3$
  5. $e = 2$, latus rectum $= 12$
A hyperbola drawn freehand is the sketch that loses marks, and this exercise is full of them. The box is the fix. Step a along the x-axis and b along the y-axis, close the rectangle, and extend the diagonals — those are the asymptotes, so the branches have somewhere to aim. The corner earns its keep twice over: it is √(a² + b²) from the centre, which is c, so the foci come off the same drawing instead of a separate formula. Both panels here use the same rectangle, one of them stood on its side.
Miscellaneous
  1. practice A conic section has eccentricity $e = 1$. Which curve is it?
    1. circle
    2. ellipse
    3. parabola
    4. hyperbola
  2. practice True or false, and correct it if false: for both an ellipse and a hyperbola, $c^2 = a^2 - b^2$.
  3. practice An ellipse and a hyperbola share $a = 5$ and $b = 3$. Find $c$ for each curve.
  4. practice Which curve results from slicing a cone with a plane parallel to its axis, cutting both nappes?
    1. circle
    2. ellipse
    3. parabola
    4. hyperbola
  5. practice Find the equation of the circle with centre $(1, 1)$ passing through the origin.
  6. practice A comet’s orbit has eccentricity $0.95$. Which curve describes its path, and why is it not a hyperbola?
Answers
  1. parabola
  2. False — that holds only for the ellipse; a hyperbola uses $c^2 = a^2 + b^2$
  3. ellipse $c = 4$; hyperbola $c = \sqrt{34}$
  4. hyperbola
  5. $(x-1)^2 + (y-1)^2 = 2$
  6. an ellipse, since $0 < 0.95 < 1$; a hyperbola needs $e > 1$
Item 3 hands both curves the same two numbers, so draw them on the same axes and the two answers separate themselves. Both cross the x-axis at 5 — that is what sharing a means. The ellipse’s focus is the green dot at 4, a unit short of that crossing; the hyperbola’s is the orange dot at 5.83, most of a unit past it. Same a, same b, one minus sign between them, and the foci end up on opposite sides of the one point both curves agree on. Item 2's answer is in the picture too: a single formula cannot put a focus in both places at once.
From the item bank
  1. A ship uses hyperbolic radio time-differences, like an old LORAN system, to fix its position: it lies on a curve where the ABSOLUTE DIFFERENCE of its distances to two fixed transmitters stays constant. Which eccentricity could this curve have?
    1. e = 0
    2. e = 1.6
    3. e = 0.6
    4. e = 1
    Check your answer
    1. e = 0 — e = 0 names a circle, which holds a single distance fixed, not an absolute difference.
    2. ✓ e = 1.6 — The absolute-difference condition is the hyperbola’s, and a hyperbola’s eccentricity is always e > 1 — 1.6 fits; the other values belong to a circle, an ellipse, or a parabola.
    3. e = 0.6 — 0 < e < 1 names an ellipse, which holds a sum of distances fixed, not an absolute difference.
    4. e = 1 — e = 1 names a parabola, which holds a point-to-line distance equal, not an absolute difference.

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Chapter-end problems

Chapter-end problems — graded set
  1. board-easy Find the equation of the circle with centre $(-2, 4)$ and radius $3$.
  2. board-easy Find the equation of the circle with centre at the origin and radius $\sqrt{7}$.
  3. board-easy Find the centre and radius of the circle $x^2 + y^2 + 8 x - 6 y - 11 = 0$.
  4. board-easy Which of these equations represents a circle of radius $5$ centred at the origin?
    1. $x^2+y^2=25$
    2. $x^2+y^2=5$
    3. $(x-5)^2+y^2=25$
    4. $x^2+(y-5)^2=5$
  5. board-easy Find the focus, directrix, and latus rectum of $y^2=16 x$.
  6. board-easy Find the equation of the parabola with vertex at the origin, opening along the positive x-axis, and focus $(5,0)$.
  7. board-easy A parabola has equation $y^2=9 x$. Find the length of its latus rectum.
  8. board-easy Find the foci, eccentricity, and latus rectum of $x^2/64+y^2/36=1$.
  9. board-easy Find the lengths of the major and minor axes of the ellipse $x^2/81+y^2/25=1$.
  10. board-easy Find the foci, eccentricity, and latus rectum of $x^2/25-y^2/144=1$.
  11. board-easy Find the equation of the hyperbola with transverse axis along the x-axis, $a=6$, and $b=8$.
  12. board-easy A conic section has eccentricity $e=0$. Which curve is it?
    1. circle
    2. ellipse
    3. parabola
    4. hyperbola
  13. board-easy Find the equation of the circle with centre $(0, -5)$ passing through the point $(3, -1)$.
  14. board-easy Find the equation of the parabola with vertex at the origin, opening along the positive x-axis, and directrix $x = -9/2$.
  15. board-standard Find the equation of the circle whose diameter has endpoints $(4, 1)$ and $(-2, 7)$.
  16. board-standard Find the centre and radius of the circle $2 x^2 + 2 y^2 - 6 x + 8 y - 12 = 0$.
  17. board-standard The latus rectum of a parabola $y^2 = 4 a x$ is $24$ units long. Find the coordinates of its focus.
  18. board-standard Find the equation of the ellipse with foci $(\pm 4, 0)$ and vertices $(\pm 7, 0)$.
  19. board-standard Find the eccentricity of the ellipse $16 x^2+25 y^2=400$.
  20. board-standard What is the latus rectum of an ellipse with $a=13$ and $b=5$?
    1. $50/13$
    2. $25/13$
    3. $13/5$
    4. $10/13$
  21. board-standard Find the equation of the hyperbola with vertices $(\pm 4, 0)$ and foci $(\pm 6, 0)$.
  22. board-standard Find the eccentricity of the hyperbola $9 x^2-16 y^2=144$.
  23. board-standard A hyperbola has $a=5$ and eccentricity $e=13/5$. What is $c$?
    1. $13$
    2. $8$
    3. $18$
    4. $5$
  24. board-standard Find the equation of the circle with centre $(5, -2)$ passing through the point $(-1, 3)$.
  25. board-standard Find the point on the parabola $y^2=12 x$ where $y=6$.
  26. board-standard Find the equation of the ellipse with major axis along the x-axis, semi-major axis $a=9$, and eccentricity $e=2/3$.
  27. board-standard Find the equation of the hyperbola with transverse axis along the x-axis, semi-transverse axis $a=8$, and eccentricity $e=5/4$.
  28. board-standard An orbit has eccentricity $e=1.2$. Which conic best describes its path?
    1. circle
    2. ellipse
    3. parabola
    4. hyperbola
  29. board-standard Find the general equation of the circle with centre $(-3, -4)$ and radius $5$.
  30. board-standard A satellite’s elliptical orbit has semi-major axis $8000$ km and eccentricity $0.05$. Find the distance between its two foci.
  31. board-standard Find the lengths of the transverse and conjugate axes of the hyperbola $x^2/49-y^2/9=1$.
  32. board-standard Find the equation of a parabola with vertex at the origin, opening along the positive x-axis, whose latus rectum equals the distance between $(0, 3)$ and $(0, -3)$.
  33. JEE Find the value of $k$ for which $x^2+y^2-4 x+6 y+k=0$ represents a circle of radius $3$.
  34. JEE An ellipse’s latus rectum equals half of its minor axis. Find its eccentricity.
  35. JEE A hyperbola’s latus rectum equals half the distance between its foci. Find its eccentricity.
  36. JEE Find the equation of the ellipse (major axis along the x-axis) that passes through the points $(4, 3)$ and $(6, 2)$.
  37. JEE Find the equation of the hyperbola with vertices $(\pm 6, 0)$ that passes through the point $(10, 16/3)$.
  38. JEE A planet’s elliptical orbit has eccentricity $e=0.5$. Its closest distance to the sun, at one focus, is $6$ units. Find its farthest distance.
  39. JEE An ellipse’s distance between foci equals its latus rectum. Find its eccentricity.
  40. JEE A circle of radius $5$ has its centre on the x-axis and passes through $(2, 4)$. Find the possible coordinates of its centre.
  41. JEE An ellipse $x^2/25+y^2/9=1$ and a hyperbola $x^2/7-y^2/b^2=1$ share the same foci. Find $b^2$.
  42. JEE If $16 x^2-9 y^2=144$, what is the eccentricity of this hyperbola?
    1. $5/3$
    2. $4/3$
    3. $5/4$
    4. $3/5$
  43. JEE An ellipse and a hyperbola share $a=5$ and $b=3$. Find the eccentricity of each curve.
Answers
  1. $(x+2)^2 + (y-4)^2 = 9$
  2. $x^2+y^2=7$
  3. centre $=(-4,3)$, radius $=6$
  4. $x^2+y^2=25$
  5. focus $=(4,0)$, directrix $x=-4$, latus rectum $=16$
  6. $y^2=20 x$
  7. $9$
  8. foci $=(\pm 2 \sqrt{7}, 0)$, $e=\sqrt{7}/4$, latus rectum $=9$
  9. major axis $=18$, minor axis $=10$
  10. foci $=(\pm 13,0)$, $e=13/5$, latus rectum $=288/5$
  11. $x^2/36-y^2/64=1$
  12. circle
  13. $x^2 + (y+5)^2 = 25$
  14. $y^2 = 18 x$
  15. $(x-1)^2 + (y-4)^2 = 18$
  16. centre $=(1.5,-2)$, radius $=3.5$
  17. focus $=(6,0)$
  18. $x^2/49+y^2/33=1$
  19. $e=3/5$
  20. $50/13$
  21. $x^2/16-y^2/20=1$
  22. $e=5/4$
  23. $13$
  24. $(x-5)^2+(y+2)^2=61$
  25. point $(3,6)$
  26. $x^2/81+y^2/45=1$
  27. $x^2/64-y^2/36=1$
  28. hyperbola
  29. $x^2+y^2+6 x+8 y=0$
  30. $800$ km
  31. transverse axis $=14$, conjugate axis $=6$
  32. $y^2=6 x$
  33. $k=4$
  34. $e=\sqrt{3}/2$
  35. $e=(1+\sqrt{17})/4$
  36. $x^2/52+y^2/13=1$
  37. $x^2/36-y^2/16=1$
  38. $18$ units
  39. $e=(\sqrt{5}-1)/2$
  40. $(-1,0)$ or $(5,0)$
  41. $b^2=9$
  42. $5/3$
  43. ellipse $e=4/5$; hyperbola $e=\sqrt{34}/5$
Problem 33 asks for one value of k and is over in a line. Turn k instead and watch what it is actually attached to. The centre never moves: it is read off the minus 4x and plus 6y terms and k is not in either of them. What k moves is the radius, through r squared equals 13 minus k — and at k equals 13 the radius reaches zero, so the circle has shrunk onto its own centre. Above that, r squared is negative and the equation describes nothing at all, which is what an exam means when it asks for what values of k this represents a circle.
Problem 38 gives you one distance and asks for the other, and the picture shows why there is nothing else it could be. The sun is at the focus, six units off centre, so it cuts the long axis into two unequal pieces: the short navy one out to the near vertex, the long purple one back to the far vertex. Those two pieces are the whole axis, so they must total 2a — and once the six and the eccentricity have given you a = 12, the 24 is fixed and the farthest distance has to be 18. Problem 30's satellite is the same drawing with different numbers: what it asks for is the gap between the two foci, 2ae.

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JEE-application problems

JEE-application problems — from the item bank
  1. A circle has centre $(3, -4)$ and radius $6$. Its general equation is $x^2 + y^2 - 6 x + 8 y + c = 0$. Find $c$.
    1. +11
    2. -11
    3. +25
    4. +19
    Check your answer
    1. +11 — $g^2+f^2-r^2 = 9+16-36 = -11$, not $11$ — check the sign of the subtraction.
    2. ✓ -11 — (B) $g=-3$, $f=4$, so $c = g^2+f^2-r^2 = 9+16-36 = -11$.
    3. +25 — $9+16=25$ stops short — the $-r^2 = -36$ term is still owed.
    4. +19 — Using $r=6$ instead of $r^2=36$ gives $9+16-6=19$, the wrong quantity entirely.
  2. The circle $x^2 + y^2 - 8 x + 2 y + 8 = 0$ has what radius?
    1. 5
    2. 4
    3. 3
    4. 9
    Check your answer
    1. 5 — $\sqrt{16+1+8}=5$ uses $+c$ — the formula needs $-c$.
    2. 4 — $\sqrt{g^2}=4$ drops both $f$ and $c$ from the radius formula.
    3. ✓ 3 — (C) $g=-4, f=1, c=8$, so radius $= \sqrt{g^2+f^2-c} = \sqrt{16+1-8} = \sqrt{9} = 3$.
    4. 9 — $16+1-8=9$ is the radicand, not the radius — take its square root.
  3. A circle has centre $(5, 12)$ and passes through the origin. Its general equation is $x^2 + y^2 - 10 x - 24 y + c = 0$. Find $c$.
    1. 169
    2. -169
    3. 13
    4. 0
    Check your answer
    1. 169 — $169$ is $g^2+f^2$, which equals $r^2$ here — but $c=g^2+f^2-r^2$, and $r^2$ cancels it exactly.
    2. -169 — $-169$ over-corrects the sign; the origin lying on the circle forces $c=0$ directly.
    3. 13 — $13$ is the radius $r$, not the constant $c$ in the general equation.
    4. ✓ 0 — (D) The circle passes through the origin, so plugging $(0,0)$ into $x^2+y^2-10x-24y+c=0$ gives $c=0$ directly — no need to compute the radius at all.
  4. A circle centred at the origin passes through the point $(6, 8)$. What is its radius?
    1. 10
    2. 14
    3. 48
    4. 100
    Check your answer
    1. ✓ 10 — (A) The radius is the distance from the origin to $(6,8)$: $r=\sqrt{6^2+8^2}=\sqrt{100}=10$.
    2. 14 — $6+8=14$ is not a distance — the radius is $\sqrt{6^2+8^2}$.
    3. 48 — $6 \times 8=48$ is not a distance either — use the distance formula.
    4. 100 — $6^2+8^2=100$ is $r^2$ — take the square root to get $r$.
  5. The circle $x^2 + y^2 + 10 x - 4 y + 13 = 0$ has what radius?
    1. $\sqrt{42}$
    2. 4
    3. 42
    4. 16
    Check your answer
    1. $\sqrt{42}$ — $\sqrt{25+4+13}=\sqrt{42}$ uses $+c$ — the formula needs $-c$.
    2. ✓ 4 — (B) $g=5, f=-2, c=13$, so radius $= \sqrt{g^2+f^2-c} = \sqrt{25+4-13} = \sqrt{16} = 4$.
    3. 42 — $25+4+13=42$ repeats the sign slip and also skips the square root.
    4. 16 — $25+4-13=16$ is the radicand, not the radius — take its square root to get $4$.
  6. A parabola has vertex at the origin and focus $(7, 0)$. What is its standard equation?
    1. $y^2 = 7 x$
    2. $y^2 = 14 x$
    3. $y^2 = 28 x$
    4. $x^2 = 28 y$
    Check your answer
    1. $y^2 = 7 x$ — $y^2=7x$ leaves out the factor of $4$ — the standard form is $y^2 = 4 a x$.
    2. $y^2 = 14 x$ — $y^2=14x$ uses $2a$ where the formula needs $4a$.
    3. ✓ $y^2 = 28 x$ — (C) Focus $(a,0)=(7,0)$ gives $a=7$, so $y^2 = 4 a x = 28 x$.
    4. $x^2 = 28 y$ — $x^2=28y$ opens upward, but the focus $(7,0)$ sits on the $x$-axis, so it opens rightward.
  7. A parabola has vertex at the origin and directrix $x = -9$. What is the value of $a$ in $y^2 = 4 a x$?
    1. -9
    2. 18
    3. 4.5
    4. 9
    Check your answer
    1. -9 — $a$ is defined as positive; the directrix is $x=-a$, so $a=9$, not $-9$.
    2. 18 — $18$ is the focus-to-directrix distance $2a$, not $a$ itself.
    3. 4.5 — $4.5$ halves $a$ instead of reading it directly off the directrix.
    4. ✓ 9 — (D) The directrix is $x=-a$, and here $x=-9$, so $a=9$.
  8. A point on the parabola $y^2 = 12 x$ has $x$-coordinate $5$. What is its distance from the focus?
    1. 8
    2. 2
    3. 6
    4. 5
    Check your answer
    1. ✓ 8 — (A) $4a=12$ gives $a=3$. The focal distance of a point on $y^2 = 4 a x$ is $x+a = 5+3 = 8$.
    2. 2 — $5-3=2$ subtracts $a$; the focal distance property is $x+a$, not $x-a$.
    3. 6 — $2a=6$ is the latus rectum’s own building block, not the focal distance.
    4. 5 — $x=5$ alone forgets the $+a$ term — the focus is not at the vertex.
  9. A point on the parabola $y^2 = 8 x$ has $y$-coordinate $12$. What is its $x$-coordinate?
    1. 144
    2. 18
    3. 9
    4. 36
    Check your answer
    1. 144 — $144$ is $y^2$, one step before the answer — still need to divide by $8$.
    2. ✓ 18 — (B) $y^2=8x$ with $y=12$: $144=8x$, so $x=144/8=18$.
    3. 9 — $9$ divides $144$ by the wrong number — check $144/8$ again.
    4. 36 — $36$ doubles the correct value — recheck the division.
  10. A parabola $y^2 = 4 a x$ has its focus and directrix a distance $10$ apart. What is the value of $a$?
    1. 10
    2. 20
    3. 5
    4. 2.5
    Check your answer
    1. 10 — $10$ is the focus-directrix distance itself, not $a$ — the distance equals $2a$, not $a$.
    2. 20 — $20$ treats the distance as $4a$; it is actually $2a$.
    3. ✓ 5 — (C) Focus $(a,0)$ and directrix $x=-a$ are $2a$ apart, so $2a=10$ gives $a=5$.
    4. 2.5 — $2.5$ divides by $4$ instead of $2$.
  11. What is the eccentricity of the ellipse $x^2/25 + y^2/9 = 1$?
    1. 3/5
    2. 5/4
    3. 9/25
    4. 4/5
    Check your answer
    1. 3/5 — $3/5$ is $b/a$, not the eccentricity $c/a$ — find $c$ first.
    2. 5/4 — $5/4$ inverts $e=c/a$ into $a/c$.
    3. 9/25 — $9/25$ is $b^2/a^2$, a different ratio entirely.
    4. ✓ 4/5 — (D) $a=5, b=3$, so $c=\sqrt{a^2-b^2}=\sqrt{25-9}=4$, and $e=c/a=4/5$.
  12. An ellipse has semi-major axis $a = 5$ and semi-minor axis $b = 3$ (major axis along $x$). Find the length of its latus rectum.
    1. 18/5
    2. 6/5
    3. 50/3
    4. 18
    Check your answer
    1. ✓ 18/5 — (A) Latus rectum $= 2b^2/a = 2(9)/5 = 18/5$.
    2. 6/5 — $2b/a=6/5$ leaves $b$ unsquared — the formula needs $2b^2/a$.
    3. 50/3 — $2a^2/b=50/3$ swaps the roles of $a$ and $b$ in the formula.
    4. 18 — $2b^2=18$ forgets the final division by $a$.
  13. An ellipse has eccentricity $e = 3/5$ and semi-major axis $a = 10$. Find $b$.
    1. 6
    2. 8
    3. 64
    4. 16
    Check your answer
    1. 6 — $c = e a = 6$ is only the halfway point — $b$ still needs $b^2=a^2-c^2$ and a square root.
    2. ✓ 8 — (B) $c = e a = (3/5)(10) = 6$. Then $b^2=a^2-c^2=100-36=64$, so $b=8$.
    3. 64 — $64$ is $b^2$, one step short — take the square root to get $b$.
    4. 16 — $16$ doubles the correct $b=8$.
  14. An ellipse has foci $(\pm 4, 0)$ and semi-major axis $a = 5$. Find $b^2$.
    1. 41
    2. 1
    3. 9
    4. 16
    Check your answer
    1. 41 — $a^2+c^2=41$ uses $+$ — the ellipse relation is $b^2=a^2-c^2$, with a minus.
    2. 1 — $a-c=1$ mixes up linear and squared quantities.
    3. ✓ 9 — (C) $c=4, a=5$, so $b^2=a^2-c^2=25-16=9$.
    4. 16 — $16$ is $c^2$ itself, not $b^2$.
  15. What are the foci of the ellipse $x^2/49 + y^2/24 = 1$?
    1. $(\pm \sqrt{73}, 0)$
    2. $(0, \pm 5)$
    3. $(\pm 25, 0)$
    4. $(\pm 5, 0)$
    Check your answer
    1. $(\pm \sqrt{73}, 0)$ — $\sqrt{a^2+b^2}=\sqrt{73}$ uses $+$ — the ellipse relation for $c$ needs a minus.
    2. $(0, \pm 5)$ — The larger denominator sits under $x^2$ here, so the major axis — and the foci — lie on the $x$-axis.
    3. $(\pm 25, 0)$ — $25$ is $a^2-b^2$, the radicand — still needs a square root to become $c$.
    4. ✓ $(\pm 5, 0)$ — (D) $a^2=49, b^2=24$, so $c=\sqrt{49-24}=\sqrt{25}=5$. Foci $=(\pm 5, 0)$.
  16. What is the eccentricity of the hyperbola $x^2/16 - y^2/9 = 1$?
    1. 5/4
    2. 4/5
    3. 3/4
    4. 5/3
    Check your answer
    1. ✓ 5/4 — (A) $a=4, b=3$, so $c=\sqrt{a^2+b^2}=\sqrt{16+9}=5$, and $e=c/a=5/4$.
    2. 4/5 — $4/5$ inverts $e=c/a$ into $a/c$.
    3. 3/4 — $3/4$ is $b/a$, not the eccentricity $c/a$.
    4. 5/3 — $5/3$ divides $c$ by $b$ instead of by $a$.
  17. A hyperbola has $a = 4$ and $b = 6$ (transverse axis along $x$). Find the length of its latus rectum.
    1. 9
    2. 18
    3. 72
    4. 16/3
    Check your answer
    1. 9 — $b^2/a=9$ leaves out the factor of $2$ — the formula is $2b^2/a$.
    2. ✓ 18 — (B) Latus rectum $= 2b^2/a = 2(36)/4 = 18$.
    3. 72 — $2b^2=72$ forgets the final division by $a$.
    4. 16/3 — $2a^2/b=16/3$ swaps the roles of $a$ and $b$.
  18. A hyperbola has foci $(\pm 13, 0)$ and $a = 5$. Find $b^2$.
    1. 194
    2. 25
    3. 144
    4. 169
    Check your answer
    1. 194 — $c^2+a^2=194$ uses $+$; the hyperbola relation for $b^2$ is $c^2-a^2$.
    2. 25 — $25$ is $a^2$ itself, not $b^2$.
    3. ✓ 144 — (C) $c=13, a=5$, so $b^2=c^2-a^2=169-25=144$.
    4. 169 — $169$ is $c^2$ itself, not $b^2$.
  19. A hyperbola has eccentricity $e = 5/3$ and $a = 6$. Find $b$.
    1. 10
    2. 64
    3. 14
    4. 8
    Check your answer
    1. 10 — $c = e a = 10$ is only the halfway point — $b$ still needs $b^2=c^2-a^2$ and a square root.
    2. 64 — $64$ is $b^2$, one step short — take the square root to get $b$.
    3. 14 — $c+a=14$ mixes a linear sum where a squared difference is needed.
    4. ✓ 8 — (D) $c = e a = (5/3)(6) = 10$. Then $b^2=c^2-a^2=100-36=64$, so $b=8$.
  20. What is the eccentricity of the hyperbola $x^2/9 - y^2/16 = 1$?
    1. 5/3
    2. 3/5
    3. 4/3
    4. 5/4
    Check your answer
    1. ✓ 5/3 — (A) $a=3, b=4$, so $c=\sqrt{a^2+b^2}=\sqrt{9+16}=5$, and $e=c/a=5/3$.
    2. 3/5 — $3/5$ inverts $e=c/a$ into $a/c$.
    3. 4/3 — $4/3$ is $b/a$, not the eccentricity $c/a$.
    4. 5/4 — $5/4$ divides $c$ by $b$ instead of by $a$.
The focal distance gives the big square and a gives the small one, so what is left over is the square on b — subtract first, then take the root, and read 12 rather than the 169 or the 25 on offer.
The last two questions in this bank look like different problems and have the same answer, which is worth a moment. One hands you an equation with 3 and 4 in it; the other hands you 6 and an eccentricity. Work both and the second turns out to be the first with every length doubled. Eccentricity does not notice, because it is a comparison between two lengths of the same curve rather than a length itself: whatever factor you apply to c you have applied to a as well, and it cancels on the way through the division. The same is true of the asymptote slope, so the two curves are not merely similar in the loose sense — they lean at exactly the same angle and would sit inside one another perfectly. This is the fastest thing you own in an exam: two hyperbolas with proportional a and b are the same shape, and any question about shape has already been answered once you have done it for either.
Twice already this chapter has said the four curves are one family. Here they are actually drawn on top of each other — same focus, same chord through it, and nothing changed between them but e. Watch the near side creep in towards the focus as e rises while the far side swings out. The green arms do not close because at e = 1 there is nothing out there to draw. In an exam this is the identification step, done before any expanding: find e, name the curve.
The prose said a ray parallel to the axis bounces through the focus. Here is the part it left out, and the part a dish depends on: the rays also arrive having travelled the same distance. The two drawn here strike at deliberately different heights and split their journeys 3.44 + 3.56 against 5 + 2 — different halves, same total. It holds for every ray, because the x subtracted on the way in is the same x added on the way out, so the total is always X₀ + a.

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