Ellipse — standard forms, semi-major - minor axes, foci, eccentricity, latus rectum: Exercises

This page drills the parent topic Ellipse. Every tool used here — the relation $c^{2} = a^{2} - b^{2}$ , eccentricity $e = c / a$ , latus rectum $2 b^{2} / a$ — was built there. We only apply them now.

A picture to keep in your head for every problem — refer back to it as each new symbol appears:

Figure — Ellipse — standard forms, semi-major - minor axes, foci, eccentricity, latus rectum

Level 1 — Recognition

Read the answer straight off the equation. No algebra beyond a square root.

Problem 1.1

For $\frac{x ^{2}}{49} + \frac{y ^{2}}{25} = 1$ , state $a$ , $b$ , and which axis is the major axis.

Recall Solution 1.1

WHAT: compare the two denominators, $49$ and $25$ . WHY: here "variable" just means the letter being squared — the $x$ in $x^{2}$ or the $y$ in $y^{2}$ . The larger denominator always sits under the major-axis variable (the letter measured along the long axis), and its square root is $a$ . In figure s01 that long axis is the horizontal one through the blue vertices. Larger is $49$ , under $x^{2}$ ⇒ $a^{2} = 49 \Rightarrow a = 7$ ; smaller $b^{2} = 25 \Rightarrow b = 5$ . Answer: $a = 7$ , $b = 5$ , major axis along the $x$ -axis (horizontal), so $x$ is the major-axis variable.

Problem 1.2

For $\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1$ , find the coordinates of the vertices and co-vertices.

Recall Solution 1.2

Larger denom $81$ is under $y^{2}$ ⇒ $a^{2} = 81 \Rightarrow a = 9$ (vertical major); $b^{2} = 16 \Rightarrow b = 4$ . Vertices lie on the major axis at $(0, \pm a) = (0, \pm 9)$ — the blue-square positions of figure s01, now turned vertical. Co-vertices lie on the minor axis at $(\pm b, 0) = (\pm 4, 0)$ — the green-dot positions, now horizontal.

Problem 1.3

Is $\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1$ an ellipse with two distinct foci? Explain.

Recall Solution 1.3

Here $a^{2} = b^{2} = 9$ , so $a = b = 3$ and $c^{2} = a^{2} - b^{2} = 0 \Rightarrow c = 0$ . Both foci sit at the centre $(0, 0)$ — they coincide (the red star in figure s01 slides all the way in to the origin). This is the degenerate/limiting case: a circle of radius $3$ , eccentricity $e = 0$ . Not two distinct foci.

Level 2 — Application

Now compute $c$ , $e$ , and the latus rectum. One relation does most of the work: $c^{2} = a^{2} - b^{2}$ .

Problem 2.1

For $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ , find $c$ , the foci, $e$ , and the latus rectum.

Recall Solution 2.1

$a^{2} = 25, b^{2} = 16$ (horizontal major, since $25 > 16$ ) ⇒ $a = 5, b = 4$ . Find $c$ . Why this formula: $c$ is the red leg of the triangle in figure s01, and Pythagoras on that triangle gives $c^{2} = a^{2} - b^{2}$ . So $c^{2} = 25 - 16 = 9 \Rightarrow c = 3$ . Foci. Why $(\pm 3, 0)$ : the foci always sit on the major axis (here the $x$ -axis) at distance $c$ from the centre — that is where the red stars live in figure s01. So $(\pm 3, 0)$ . Eccentricity. Why $c / a$ : $e$ measures how far the focus has travelled toward the vertex as a fraction of the full semi-major length; that fraction is exactly $c / a$ . So $e = 3/5 = 0.6$ . Latus rectum. Why $2 b^{2} / a$ : it is the vertical chord's full length where the ellipse crosses the focus; the parent note derived it as $2 b^{2} / a$ . So $LR = 2 (16) /5 = 32/5 = 6.4$ .

Problem 2.2

Normalise and analyse $4 x^{2} + 9 y^{2} = 36$ : find $a, b, c, e$ and the foci.

Recall Solution 2.2

WHAT: divide through by $36$ so the right side is $1$ . WHY: every formula ( $a^{2}$ = denominator, $c^{2} = a^{2} - b^{2}$ , etc.) assumes the standard form, which is only readable when RHS $= 1$ . $\frac{4 x ^{2}}{36} + \frac{9 y ^{2}}{36} = 1 \Rightarrow \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ . Larger denom $9$ under $x^{2}$ ⇒ $a^{2} = 9, b^{2} = 4$ ⇒ $a = 3, b = 2$ (horizontal major). Find $c$ . Why $c^{2} = a^{2} - b^{2}$ : again the Pythagoras triangle of figure s01 — knowing $a$ and $b$ gives the red leg $c$ . So $c^{2} = 9 - 4 = 5 \Rightarrow c = 5 \approx 2.236$ . Foci. Why on the $x$ -axis: the major axis is horizontal, and foci always lie on it. So $(\pm 5, 0)$ . Eccentricity. Why $c / a$ : the fraction of the semi-major length occupied by $c$ . So $e = 5 /3 \approx 0.745$ .

Problem 2.3

For $\frac{x ^{2}}{4} + \frac{y ^{2}}{36} = 1$ , find the eccentricity and the length of the latus rectum.

Recall Solution 2.3

Larger denom $36$ under $y^{2}$ ⇒ $a^{2} = 36, b^{2} = 4$ ⇒ $a = 6, b = 2$ (vertical major — imagine figure s01 stood up on its end). Find $c$ . Why $c^{2} = a^{2} - b^{2}$ : the same triangle, now vertical; two of its sides are known. So $c^{2} = 36 - 4 = 32 \Rightarrow c = 32 = 42 \approx 5.657$ . Eccentricity. Why $c / a$ (not $c / b$ ): $e$ is always measured against the semi-major length $a$ , since that is the ellipse's own scale. So $e = 42 /6 = 22 /3 \approx 0.943$ . Latus rectum. Why $2 b^{2} / a$ : the through-focus chord formula holds in either orientation, using semi-minor $b$ over semi-major $a$ . So $LR = 2 (4) /6 = 8/6 = 4/3 \approx 1.333$ .

Level 3 — Analysis

Reverse the process, or reason about how a parameter changes the shape.

Problem 3.1

Build the ellipse: foci at $(0, \pm 4)$ and eccentricity $e = \frac{4}{5}$ . Find its equation.

Recall Solution 3.1

Foci on the $y$ -axis ⇒ vertical major, and $c = 4$ (the focus is at distance $c$ from centre). Find $a$ . Why $a = c / e$ : rearranging the definition $e = c / a$ isolates $a$ ; we know $c$ and $e$ , so $a = 4/ (4/5) = 5$ , hence $a^{2} = 25$ . Find $b^{2}$ . Why $b^{2} = a^{2} - c^{2}$ : the Pythagoras triangle again — the green leg $b$ from the other two sides. So $b^{2} = 25 - 16 = 9$ . Vertical major means $a^{2}$ goes under $y^{2}$ : $\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1.$

Problem 3.2

An ellipse has vertices $(\pm 6, 0)$ and passes through $(3, 23)$ . Find $b^{2}$ and the eccentricity.

Recall Solution 3.2

Vertices $(\pm 6, 0)$ ⇒ horizontal major, $a = 6$ , $a^{2} = 36$ . Equation so far: $\frac{x ^{2}}{36} + \frac{y ^{2}}{b ^{2}} = 1$ . WHAT: substitute the point $(3, 23)$ . WHY: a point on the curve must satisfy the equation — that's one equation in the one unknown $b^{2}$ . $\frac{9}{36} + \frac{( 2 3 ) ^{2}}{b ^{2}} = 1 \Rightarrow \frac{1}{4} + \frac{12}{b ^{2}} = 1 \Rightarrow \frac{12}{b ^{2}} = \frac{3}{4} \Rightarrow b^{2} = 16$ . Check $b^{2} < a^{2}$ : $16 < 36$ ✓ (consistent with horizontal major). Find $c, e$ . Why $c^{2} = a^{2} - b^{2}$ : the triangle of figure s01 with $a, b$ known. So $c^{2} = 36 - 16 = 20 \Rightarrow c = 25$ ; and why $e = c / a$ : the fraction of the semi-major length, giving $e = 25 /6 = 5 /3 \approx 0.745$ .

Problem 3.3

Two ellipses share the same $a = 10$ . Ellipse A has $c = 6$ ; ellipse B has $c = 8$ . Which is more squashed, and by how much do their latus recta differ?

Recall Solution 3.3

Ellipse A: $b^{2} = a^{2} - c^{2} = 100 - 36 = 64$ so $b_{A} = 8$ ; $e_{A} = 6/10 = 0.6$ ; $LR_{A} = 2 (64) /10 = 12.8$ . Ellipse B: $b^{2} = 100 - 64 = 36$ so $b_{B} = 6$ ; $e_{B} = 8/10 = 0.8$ ; $LR_{B} = 2 (36) /10 = 7.2$ . Why larger $e$ means more squashed — read it off figure s02. Both curves have the same half-width $a = 10$ (they reach the same left/right vertices). But look at the heights: since $b = a^{2} - c^{2}$ , pushing $c$ up (as B does) drains $b$ down — B's green top sits lower than A's. A shorter $b$ with an unchanged $a$ is precisely a flatter, more squashed oval. In the figure, B (orange) is visibly thinner and hugs the horizontal axis, while A (green) is rounder. So B is more squashed, and its foci (orange stars) sit closer to the vertices. Latus recta. The dashed vertical chords in figure s02 are the latus recta; the thinner ellipse B has the shorter chord. Difference: $12.8 - 7.2 = 5.6$ .

Level 4 — Synthesis

Combine several facts, or translate a physical/word description into the algebra.

Problem 4.1

The distance between the foci of an ellipse is $8$ , and the length of the minor axis is $6$ . Find the equation (horizontal major) and the eccentricity.

Recall Solution 4.1

Distance between foci $= 2 c = 8 \Rightarrow c = 4$ . Minor axis length $= 2 b = 6 \Rightarrow b = 3$ , so $b^{2} = 9$ . Find $a$ . Why $a^{2} = b^{2} + c^{2}$ : this is the same Pythagoras triangle of figure s01 read the other way — hypotenuse $a$ from legs $b, c$ . So $a^{2} = 9 + 16 = 25 \Rightarrow a = 5$ . Horizontal major ⇒ $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . $e = c / a = 4/5 = 0.8$ .

Problem 4.2

An ellipse has eccentricity $e = \frac{1}{2}$ and latus rectum of length $6$ . Find $a$ and $b$ (assume horizontal major).

Recall Solution 4.2

Two facts, two unknowns. Write both in terms of $a, b$ . $LR = \frac{2 b ^{2}}{a} = 6 \Rightarrow b^{2} = 3 a$ . …(i) $e = \frac{c}{a} = \frac{1}{2} \Rightarrow c = \frac{a}{2} \Rightarrow c^{2} = \frac{a ^{2}}{4}$ . Use $b^{2} = a^{2} - c^{2}$ : $b^{2} = a^{2} - \frac{a ^{2}}{4} = \frac{3 a ^{2}}{4}$ . …(ii) Set (i) $=$ (ii): $3 a = \frac{3 a ^{2}}{4} \Rightarrow 1 = \frac{a}{4} \Rightarrow a = 4$ . Then $b^{2} = 3 a = 12$ , so $b = 23 \approx 3.464$ . Answer: $a = 4$ , $b^{2} = 12$ ; equation $\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1$ .

Problem 4.3

A whispering gallery has an elliptical cross-section $10$ m wide (full major axis) and $6$ m tall (full minor axis). Two people stand at the foci. How far apart are they?

Recall Solution 4.3

Full major axis $= 2 a = 10 \Rightarrow a = 5$ ; full minor axis $= 2 b = 6 \Rightarrow b = 3$ . Distance between foci $= 2 c$ , and why $c^{2} = a^{2} - b^{2}$ : the triangle of figure s01 with $a, b$ known. So $c^{2} = 25 - 9 = 16 \Rightarrow c = 4$ . They stand $2 c = 8$ m apart. (Physically: a sound from one focus reflects off the wall and reconverges at the other focus — that's the reflective property behind whispering galleries.)

Level 5 — Mastery

Prove a general fact, or push a limiting case.

Problem 5.1

Prove that for any ellipse, the latus rectum equals $2 a (1 - e^{2})$ .

Recall Solution 5.1

Start from what's given: $LR = \frac{2 b ^{2}}{a}$ and $b^{2} = a^{2} - c^{2}$ . Substitute the second into the first: $LR = \frac{2 ( a ^{2} - c ^{2} )}{a}$ . Now factor out $a^{2}$ inside the bracket. Why: we want the ratio $c^{2} / a^{2}$ to appear, because that ratio is $e^{2}$ — that's the whole point of the proof. Divide and multiply by $a^{2}$ : $\frac{2 ( a ^{2} - c ^{2} )}{a} = \frac{2 a ^{2}}{a} \cdot \frac{a ^{2} - c ^{2}}{a ^{2}} = 2 a (\frac{a ^{2}}{a ^{2}} - \frac{c ^{2}}{a ^{2}}) = 2 a (1 - \frac{c ^{2}}{a ^{2}}) .$ Replace the ratio. Why: $\frac{c ^{2}}{a ^{2}} = (\frac{c}{a})^{2} = e^{2}$ by the definition $e = c / a$ . Therefore $LR = 2 a (1 - e^{2})$ . $■$ Sanity check the extremes: as $e \to 0$ (circle), $LR \to 2 a$ — the full diameter, correct for a circle. As $e \to 1$ (flat), $LR \to 0$ — the ellipse collapses to a segment. Both limits make sense.

Problem 5.2

Show that as $e \to 1^{-}$ with $a$ held fixed, the ellipse flattens: prove $b \to 0$ . What does the shape approach?

Recall Solution 5.2

From $e = c / a$ we get $c = e a$ , so $c^{2} = e^{2} a^{2}$ . Then $b^{2} = a^{2} - c^{2} = a^{2} - e^{2} a^{2} = a^{2} (1 - e^{2})$ , giving $b = a 1 - e^{2}$ . As $e \to 1^{-}$ , $(1 - e^{2}) \to 0^{+}$ , so $b \to 0$ . Shape: the minor half-axis vanishes while the major stays $a$ ; the ellipse squashes onto the segment from $(- a, 0)$ to $(a, 0)$ — a degenerate "line-segment ellipse." The foci $c = e a \to a$ race out to the vertices. This is the boundary beyond which the conic would open up into a parabola ( $e = 1$ ) and then a hyperbola ( $e > 1$ ).

Problem 5.3

A point $P$ on the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ has one focal distance $P F_{1} = 7$ . Find the other focal distance $P F_{2}$ and verify with the defining property.

Recall Solution 5.3

$a^{2} = 25 \Rightarrow a = 5$ , so the defining constant is $2 a = 10$ . Defining property: $P F_{1} + P F_{2} = 2 a = 10$ . Given $P F_{1} = 7$ ⇒ $P F_{2} = 10 - 7 = 3$ . Why each focal distance must lie in $[a - c, a + c]$ . As $P$ travels around the ellipse, its distance to a focus is largest when $P$ is at the far vertex and smallest at the near vertex — the focus sits on the major axis at distance $c$ from centre, and the vertices are at distance $a$ from centre, so the two extreme focal distances are $a + c$ (far vertex) and $a - c$ (near vertex). Every other position gives something in between. Here $c^{2} = 25 - 9 = 16 \Rightarrow c = 4$ , so the allowed range is $[a - c, a + c] = [1, 9]$ . Check both values. $P F_{1} = 7 \in [1, 9]$ and $P F_{2} = 3 \in [1, 9]$ ✓ — so this configuration is geometrically possible.

Recall One-glance summary of every tool used

Bigger denominator ⇒ $a^{2}$ ⇒ major axis direction ⇒ where the foci live.
Triangle relation: $a^{2} = b^{2} + c^{2}$ (equivalently $c^{2} = a^{2} - b^{2}$ , $b^{2} = a^{2} - c^{2}$ ).
$e = c / a \in [0, 1)$ ; $b = a 1 - e^{2}$ ; $LR = \frac{2 b ^{2}}{a} = 2 a (1 - e^{2})$ .
Full lengths: major $= 2 a$ , minor $= 2 b$ , focal separation $= 2 c$ ; focal-sum $= 2 a$ .
Each focal distance ranges over $[a - c, a + c]$ .

Connections

Parent: Ellipse (Hinglish)
Conic Sections — overview
Circle (the $e = 0$ limit in Problem 1.3)
Parabola and Hyperbola — standard forms (what lies past $e = 1$ , Problem 5.2)
Eccentricity and directrix
Pythagorean theorem (the $a^{2} = b^{2} + c^{2}$ triangle in figure s01)