MathsISC Class 11 · Conic Sections

Ellipse and Hyperbola: Standard Equations, Foci, Eccentricity and Latus Rectum

Updated 17 October 2026 · Fact-checked

The ellipse x²/a² + y²/b² = 1 (a > b) has foci (±c, 0) with c² = a² − b² and eccentricity e = c/a < 1. The hyperbola x²/a² − y²/b² = 1 has foci (±c, 0) with c² = a² + b², e = c/a > 1 and asymptotes y = ±(b/a)x. In both, the latus rectum has length 2b²/a.

Understand Ellipse and Hyperbola: Standard Forms

Ellipse. The locus of a point whose distances from two fixed points (foci) add to a constant 2a (greater than the distance between the foci), or the conic with $e < 1$ . For $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with $a > b$ : centre at the origin; major axis along the x-axis, length 2a; minor axis length 2b; vertices $(\pm a, 0)$ ; foci $(\pm c, 0)$ with $c^{2} = a^{2} - b^{2}$ ; $e = \frac{c}{a}$ ; directrices $x = \pm \frac{a}{e}$ ; latus rectum $\frac{2 b ^{2}}{a}$ . If $b > a$ the major axis lies along the y-axis and $c^{2} = b^{2} - a^{2}$ .

Hyperbola. The locus of a point whose distances from two foci differ by a constant 2a, or the conic with $e > 1$ . For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : transverse axis along the x-axis, length 2a; conjugate axis length 2b; vertices $(\pm a, 0)$ ; foci $(\pm c, 0)$ with $c^{2} = a^{2} + b^{2}$ ; $e = \frac{c}{a} > 1$ ; latus rectum $\frac{2 b ^{2}}{a}$ ; asymptotes $y = \pm \frac{b}{a} x$ . For $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ the foci are on the y-axis at $(0, \pm c)$ .

Reading an equation. Divide to get 1 on the right side; compare the denominators. In an ellipse the larger denominator belongs to the major axis. In a hyperbola the positive term decides the transverse axis.

Finding the equation. Use the given foci to get c, use e or the axis length to find a, then get $b^{2}$ from $c^{2} = a^{2} - b^{2}$ (ellipse) or $c^{2} = a^{2} + b^{2}$ (hyperbola).

Comparison. Ellipse: bounded, sum of focal distances constant, c < a. Hyperbola: two branches, difference of focal distances constant, c > a.

Ellipse and Hyperbola: Standard Forms: formulas to remember

Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $c^2 = a^2 - b^2$, $e = \frac{c}{a}$; Latus rectum $\frac{2 b ^{2}}{a}$ .
Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $c^2 = a^2 + b^2$, $e = \frac{c}{a}$; Asymptotes $y = \pm \frac{b}{a} x$ .
Eccentricity: $Ellipse: $0 < e < 1$ | Parabola: $e = 1$ | Hyperbola: $e > 1$$; Focus-directrix ratio.
Focal property: $Ellipse: $PF_1 + PF_2 = 2a$ | Hyperbola: $|PF_1 - PF_2| = 2a$$; Defines the curves.

Ellipse and Hyperbola: Standard Forms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

An elliptical garden path

Suppose a garden is the ellipse $4 x^{2} + 9 y^{2} = 36$ (metres).

Find its axes, foci and eccentricity.

Try it first, then show the solutionSolution

$\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ : $a = 3$ , $b = 2$ .
Major axis $= 2 a = 6$ m, minor axis $= 2 b = 4$ m.
$c = 9 - 4 = 5$ : foci $(\pm 5, 0)$ , $e = \frac{5}{3} \approx 0.745$ .

Answer: Axes 6 m and 4 m; foci (±√5, 0); e = √5/3.

Memory hook: The larger denominator is a².

Tricks to remember Ellipse and Hyperbola: Standard Forms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Ellipse: minus
c² = a² − b² for the ellipse.
Use it when: Remembering the relation.
Hyperbola: plus
c² = a² + b² for the hyperbola, and c > a.
Use it when: Remembering the relation.

Ellipse and Hyperbola: Standard Forms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Eccentricity of an ellipse · 1 mark · MCQBoard level

For the ellipse $9 x^{2} + 25 y^{2} = 225$ the eccentricity is:
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{4}{3}$
(d) $\frac{5}{4}$

Try it first, then show the solutionSolution

$\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ : $a = 5$ , $b = 3$ , $c = 4$ , $e = \frac{4}{5}$ .

Answer: (b) $\frac{4}{5}$

Example 2: Ellipse elements · 3 marks · Short answerBoard level

Find the lengths of the axes, the foci, the eccentricity and the latus rectum of $4 x^{2} + 9 y^{2} = 36$ .

Try it first, then show the solutionSolution

$\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ : $a = 3$ , $b = 2$ ; axes 6 and 4. (1 mark)
$c = 5$ ; foci $(\pm 5, 0)$ ; $e = \frac{5}{3}$ . (1 mark)
Latus rectum $= \frac{2 b ^{2}}{a} = \frac{8}{3}$ . (1 mark)

Answer: Axes 6, 4; foci (±√5, 0); e = √5/3; LR = 8/3.

Example 3: Ellipse from foci and e · 3 marks · Short answerHarder board level

Find the equation of the ellipse with foci $(\pm 3, 0)$ and eccentricity $\frac{3}{5}$ .

Try it first, then show the solutionSolution

$c = 3$ and $e = \frac{c}{a} = \frac{3}{5}$ , so $a = 5$ . (1 mark)
$b^{2} = a^{2} - c^{2} = 25 - 9 = 16$ . (1 mark)
Equation: $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ . (1 mark)

Answer: x²/25 + y²/16 = 1

Example 4: Hyperbola elements · 4 marks · Short answerHarder board level

Find the foci, eccentricity, asymptotes and latus rectum of $16 x^{2} - 9 y^{2} = 144$ .

Try it first, then show the solutionSolution

$\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ : $a = 3$ , $b = 4$ . (1 mark)
$c = 9 + 16 = 5$ ; foci $(\pm 5, 0)$ ; $e = \frac{5}{3}$ . (1.5 marks)
Asymptotes $y = \pm \frac{4}{3} x$ . (0.75 mark)
Latus rectum $= \frac{2 \times 16}{3} = \frac{32}{3}$ . (0.75 mark)

Answer: Foci (±5, 0); e = 5/3; y = ±4x/3; LR = 32/3.

Example 5: Equation of a hyperbola from data · 5 marks · Long answerHOTS / competency

Find the equation of the hyperbola with foci $(0, \pm 5)$ and transverse axis of length 8. Find its eccentricity and compare with the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ .

Try it first, then show the solutionSolution

Foci on the y-axis, $c = 5$ ; transverse axis $2 a = 8$ , so $a = 4$ . (1.5 marks)
$b^{2} = c^{2} - a^{2} = 25 - 16 = 9$ . Equation: $\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1$ . (1.5 marks)
Eccentricity $e = \frac{5}{4} > 1$ . (1 mark)
For the ellipse $a = 4$ , $b = 3$ , $c = 7$ , $e = \frac{7}{4} < 1$ : the same a and b but different signs and relation. (1 mark)

Answer: (y²/16) − (x²/9) = 1; e = 5/4 (the ellipse has e = √7/4).

Where students lose marks in Ellipse and Hyperbola: Standard Forms

The slips examiners see most often, and what to write instead.

Mistake: Using + for the ellipse relation
Writing c² = a² + b².
Do this: Ellipse: c² = a² − b². Hyperbola: c² = a² + b².
Usually costs: 2 marks.
Mistake: Wrong axis assignment
Taking the first denominator as a² without checking.
Do this: In an ellipse a² is the larger denominator.
Usually costs: 1 to 2 marks.
Mistake: Forgetting to reduce to standard form
Reading a, b from 4x² + 9y² = 36.
Do this: Divide by 36 first.
Usually costs: 1 mark.

Exam tips

Reduce to the form "= 1" first.
Identify a, b, c before finding anything else.
Label foci and vertices on a sketch.
Verify that e < 1 for an ellipse and e > 1 for a hyperbola.

Practice questions on Ellipse and Hyperbola: Standard Forms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The foci of the hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ are:
(a) $(\pm 4, 0)$
(b) $(\pm 5, 0)$
(c) $(\pm 7, 0)$
(d) $(0, \pm 5)$
Hint
$c^{2} = 16 + 9$ .
Answer
(b) $(\pm 5, 0)$
Q2 · 2 marks · Short answerBoard level
Find the equation of the ellipse with major axis 10 and minor axis 8, centre at the origin and major axis along x.
Hint
a = 5, b = 4.
Answer
$\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ .
Q3 · 3 marks · Short answerHarder board level
Find the eccentricity and foci of $25 x^{2} + 9 y^{2} = 225$ .
Hint
Major axis along y.
Answer
$\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1$ : $b > a$ , major along y; $c^{2} = 25 - 9 = 16$ ; foci $(0, \pm 4)$ ; $e = \frac{4}{5}$ .
Q4 · 3 marks · Short answerHarder board level
Find the equation of the hyperbola with vertices $(\pm 3, 0)$ and eccentricity $\frac{5}{3}$ .
Hint
a = 3, c = 5.
Answer
$b^{2} = 25 - 9 = 16$ ; $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ .
Q5 · 3 marks · Short answerHarder board level
Show that the point (4, 0) lies on $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ and find the sum of its focal distances.
Hint
Sum equals 2a.
Answer
$\frac{16}{16} = 1$ , so (4, 0) lies on the curve (a vertex). The sum of its focal distances equals $2 a = 8$ (indeed distances $4 - 7$ and $4 + 7$ ).
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a whispering gallery has an elliptical ceiling with major axis 40 m and minor axis 24 m.
(i) Find a and b. (1 mark)
(ii) Find the distance between the foci. (1 mark)
(iii) Find the eccentricity. (1 mark)
(iv) Write the equation with the centre at the origin. (1 mark)
Hint
Foci are at ±c.
Answer
(i) $a = 20$ , $b = 12$ . (ii) $c = 400 - 144 = 16$ ; distance between foci 32 m. (iii) $e = \frac{16}{20} = 0.8$ . (iv) $\frac{x ^{2}}{400} + \frac{y ^{2}}{144} = 1$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.c² for an ellipse?Show answer
a² − b².
2.c² for a hyperbola?Show answer
a² + b².
3.Asymptotes?Show answer
y = ±(b/a)x.
4.LR length?Show answer
2b²/a.

Ellipse and Hyperbola: Standard Forms: frequently asked questions

What is the condition a > b for?

It makes the major axis lie along the x-axis; otherwise the roles of x and y are interchanged.

What are asymptotes?

Straight lines that the branches of the hyperbola approach without meeting.

What does eccentricity measure?

The shape: near 0 for a nearly circular ellipse, near 1 for a long thin ellipse.