MathsCBSE Class 11 · Conic Sections

Ellipse and Hyperbola: Foci, Eccentricity and Equations (Class 11)

Q: How do I tell an ellipse from a hyperbola?

An ellipse has a plus sign between the x² and y² terms; a hyperbola has a minus sign.

Q: Can b be larger than a in an ellipse?

Yes, but then the major axis is along the y-axis and the roles of a and b swap in the formulas for c and e.

Q: What is the hyperbola's transverse axis?

The line segment joining the two vertices; its length is 2a.

Q: Is the circle a special ellipse?

Yes: a = b gives c = 0 and e = 0. The 2026-27 syllabus has removed only the NCERT section on special cases of the ellipse, so the main treatment stays.

Updated 17 October 2026 · Fact-checked

For the ellipse x²/a² + y²/b² = 1 (a > b), c² = a² − b², foci (±c, 0), e = c/a < 1 and latus rectum 2b²/a. For the hyperbola x²/a² − y²/b² = 1, c² = a² + b², foci (±c, 0), e = c/a > 1 and latus rectum 2b²/a. Sum of focal distances on an ellipse is 2a; difference on a hyperbola is 2a.

Red dots are the foci: on the ellipse the two focal distances add to 2a, on the hyperbola they differ by 2a, and c² = a² − b² versus a² + b².

Illustration drawn for QuestionBuddy.

Understand Ellipse and Hyperbola: Foci, Eccentricity and Equations

Ellipse. The set of points whose distances from two fixed points (foci) have a constant sum $2 a$ , with $2 a$ greater than the distance between the foci. Standard form (centre at origin, major axis along the $x$ -axis):
$ $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b > 0, c^{2} = a^{2} - b^{2} .$ $
Vertices $(\pm a, 0)$ , foci $(\pm c, 0)$ , major axis length $2 a$ , minor axis length $2 b$ , eccentricity $e = \frac{c}{a} \in (0, 1)$ , latus rectum $\frac{2 b ^{2}}{a}$ . If the larger denominator sits under $y^{2}$ , the major axis is along the $y$ -axis and foci are $(0, \pm c)$ . Always compare the denominators to decide which axis is major.

Hyperbola. The set of points whose distances from two foci differ by a constant $2 a$ , less than the distance between the foci. Standard form (transverse axis along the $x$ -axis):
$ $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, c^{2} = a^{2} + b^{2} .$ $
Vertices $(\pm a, 0)$ , foci $(\pm c, 0)$ , $e = \frac{c}{a} > 1$ , latus rectum $\frac{2 b ^{2}}{a}$ . For $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ the transverse axis is along $y$ and the foci are $(0, \pm c)$ . Here the transverse axis is determined by the positive term, and $a$ is the denominator of that term (it need not be larger than $b$ ). A hyperbola with $a = b$ is rectangular and $e = 2$ .

Focal distances. On an ellipse, a point $(x_{1}, y_{1})$ has focal distances $a \pm e x_{1}$ whose sum is $2 a$ . On a hyperbola the focal distances are $∣ e x_{1} \pm a ∣$ whose difference is $2 a$ .

Building the equation. Identify the conic type and the orientation; use the data to find $a$ , $b$ or $c$ :

foci and major axis (or vertices): $a$ and $c$ give $b$ via the relation above;
eccentricity with a focus or vertex: $c = e a$ ;
a point on the curve: substitute to get one equation in $a^{2}$ , $b^{2}$ ; two points give two linear equations in $\frac{1}{a ^{2}}$ and $\frac{1}{b ^{2}}$ .

Comparison. Ellipse: sum constant, $c < a$ , $e < 1$ . Hyperbola: difference constant, $c > a$ , $e > 1$ . Parabola: $e = 1$ .

Ellipse and Hyperbola: Foci, Eccentricity and Equations: formulas to remember

Ellipse: $x^{2} / a^{2} + y^{2} / b^{2} = 1; c^{2} = a^{2} - b^{2}; e = c / a$; Foci (±c, 0) when a > b.
Hyperbola: $x^{2} / a^{2} - y^{2} / b^{2} = 1; c^{2} = a^{2} + b^{2}; e = c / a$; Foci (±c, 0).
Latus rectum: $2 b^{2} / a f or b o t h e l l i p se an d h y p er b o l a$; Chord through a focus, perpendicular to the major or transverse axis.
Focal distances (ellipse): $a + e x 1 an d a - e x 1; s u m 2 a$; For the point (x1, y1).
Rectangular hyperbola: $x^{2} - y^{2} = a^{2}; e = \sqrt2$; Because b = a.

How to solve Ellipse and Hyperbola: Foci, Eccentricity and Equations questions

Use these steps for reading and building equations of an ellipse or a hyperbola.

1Write the equation in the standard form with $1$ on the right-hand side.
2For an ellipse, the larger denominator is $a^{2}$ ; for a hyperbola, $a^{2}$ is the denominator of the positive term.
3Find $c$ from $c^{2} = a^{2} - b^{2}$ (ellipse) or $c^{2} = a^{2} + b^{2}$ (hyperbola).
4Write the vertices, foci, eccentricity $e = c / a$ and latus rectum $2 b^{2} / a$ with the correct axis.
5For an equation, use the data to find $a^{2}$ and $b^{2}$ and place them according to the orientation.
6Verify by checking the definition (sum or difference of focal distances) at a point.

Ellipse and Hyperbola: Foci, Eccentricity and Equations 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.

Foci of an elliptical hall

Suppose a whispering gallery has an elliptical floor 20 m long and 12 m wide. A whisper at one focus is heard clearly at the other.

How far apart should two people stand at the foci?

Try it first, then show the solutionSolution

Major axis $2 a = 20$ , so $a = 10$ ; minor axis $2 b = 12$ , so $b = 6$ .
$c = a^{2} - b^{2} = 100 - 36 = 8$ .
The foci are $(\pm 8, 0)$ , so the distance between them is $2 c = 16$ m.
Eccentricity $e = \frac{8}{10} = 0.8$ .

Answer: 16 m apart.

Memory hook: a, b, c form a right triangle for the ellipse with a as the hypotenuse.

Hyperbola from a time difference

Suppose two radio stations sit 10 km apart on a straight east-west line, with the origin midway. A ship measures that it is 6 km nearer to one station than to the other.

On what curve must the ship lie? Write its equation.

Try it first, then show the solutionSolution

The difference of distances is constant, $2 a = 6$ , so $a = 3$ .
The stations are the foci: $2 c = 10$ , $c = 5$ .
$b^{2} = c^{2} - a^{2} = 25 - 9 = 16$ .
Curve: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ (the branch nearer the closer station).

Answer: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$

Memory hook: Constant difference means hyperbola; constant sum means ellipse.

Tricks to remember Ellipse and Hyperbola: Foci, Eccentricity and Equations

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

Ellipse subtracts, hyperbola adds
Ellipse: c² = a² − b². Hyperbola: c² = a² + b² (the hyperbola has the bigger c).
Use it when: Never mix the two relations.
Bigger denominator, bigger axis
For an ellipse, the larger denominator marks the major axis.
Use it when: Deciding the foci axis.
Positive term for hyperbola
The variable with the positive sign carries the transverse axis.
Use it when: Deciding the hyperbola orientation.
Same latus rectum formula
2b²/a for both, so you only learn one formula.
Use it when: LR questions.

Ellipse and Hyperbola: Foci, Eccentricity and Equations: solved examples

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

Example 1: Reading an ellipse · 2 marks · Very short answerBoard level

Find the foci, vertices, eccentricity and latus rectum of $x^{2} + 4 y^{2} = 16$ .

Try it first, then show the solutionSolution

Divide by 16: $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ , so $a = 4$ , $b = 2$ . (½ mark)
$c = 16 - 4 = 23$ ; foci $(\pm 23, 0)$ ; vertices $(\pm 4, 0)$ . (½ mark)
$e = \frac{2 3}{4} = \frac{3}{2}$ . (½ mark)
Latus rectum $= \frac{2 \cdot 4}{4} = 2$ . (½ mark)

Answer: Foci $(\pm 23, 0)$ ; vertices $(\pm 4, 0)$ ; $e = \frac{3}{2}$ ; LR $= 2$

Example 2: Reading a hyperbola · 3 marks · Short answerBoard level

Find the foci, vertices, eccentricity and latus rectum of $x^{2} - 4 y^{2} = 16$ .

Try it first, then show the solutionSolution

Divide by 16: $\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1$ , so $a = 4$ , $b = 2$ . (1 mark)
$c = 16 + 4 = 25$ ; foci $(\pm 25, 0)$ ; vertices $(\pm 4, 0)$ . (1 mark)
$e = \frac{2 5}{4} = \frac{5}{2}$ ; latus rectum $= \frac{2 \cdot 4}{4} = 2$ . (1 mark)

Answer: Foci $(\pm 25, 0)$ ; vertices $(\pm 4, 0)$ ; $e = \frac{5}{2}$ ; LR $= 2$

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

Find the equation of the ellipse with centre at the origin, foci on the $x$ -axis, eccentricity $\frac{2}{3}$ and latus rectum of length 10.

Try it first, then show the solutionSolution

$c = \frac{2 a}{3}$ , so $b^{2} = a^{2} - c^{2} = \frac{5 a ^{2}}{9}$ . (1 mark)
Latus rectum: $\frac{2 b ^{2}}{a} = \frac{10 a}{9} = 10$ , so $a = 9$ and $b^{2} = 45$ . (1 mark)
Equation $\frac{x ^{2}}{81} + \frac{y ^{2}}{45} = 1$ . Check: $c = 6$ , $e = \frac{6}{9} = \frac{2}{3}$ , $\frac{2 \cdot 45}{9} = 10$ . (1 mark)

Answer: $\frac{x ^{2}}{81} + \frac{y ^{2}}{45} = 1$

Example 4: Hyperbola from vertices and eccentricity · 3 marks · Short answerHarder board level

Find the equation of the hyperbola with vertices $(0, \pm 4)$ and eccentricity $\frac{3}{2}$ . Also give its foci.

Try it first, then show the solutionSolution

Vertices on the $y$ -axis, so $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ with $a = 4$ . (½ mark)
$c = e a = \frac{3}{2} \cdot 4 = 6$ . (1 mark)
$b^{2} = c^{2} - a^{2} = 36 - 16 = 20$ . (1 mark)
Equation $\frac{y ^{2}}{16} - \frac{x ^{2}}{20} = 1$ ; foci $(0, \pm 6)$ . (½ mark)

Answer: $\frac{y ^{2}}{16} - \frac{x ^{2}}{20} = 1$ ; foci $(0, \pm 6)$

Example 5: Focal distances on an ellipse · 5 marks · Long answerHOTS / competency

Show that for the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ the sum of the focal distances of the point $P (3, \frac{12}{5})$ is $2 a$ , and verify that they equal $a \pm e x$ .

Try it first, then show the solutionSolution

Check $P$ lies on the curve: $\frac{9}{25} + \frac{144/25}{9} = \frac{9}{25} + \frac{16}{25} = 1$ . (1 mark)
$a = 5$ , $b = 3$ , $c = 4$ , $e = \frac{4}{5}$ ; foci $(\pm 4, 0)$ . (1 mark)
$P F_{1} = (3 - 4)^{2} + \frac{144}{25} = \frac{169}{25} = \frac{13}{5}$ . (1 mark)
$P F_{2} = 49 + \frac{144}{25} = \frac{1369}{25} = \frac{37}{5}$ . (1 mark)
Sum $= \frac{50}{5} = 10 = 2 a$ . Also $a - e x = 5 - \frac{4}{5} \cdot 3 = \frac{13}{5}$ and $a + e x = \frac{37}{5}$ . (1 mark)

Answer: $P F_{1} = \frac{13}{5}$ , $P F_{2} = \frac{37}{5}$ , sum $= 10 = 2 a$

Where students lose marks in Ellipse and Hyperbola: Foci, Eccentricity and Equations

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

Mistake: Using c² = a² − b² for a hyperbola
The two relations are mixed up.
Wrong working
$\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ : $c = 9 - 16$ .
Right working
$c = 9 + 16 = 5$ .
Do this: Hyperbola adds; remember c is the largest.
Usually costs: Foci, eccentricity and LR.
Mistake: Taking the larger denominator as a² for a hyperbola
That rule is for an ellipse only.
Wrong working
$\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ read as $a^{2} = 16$ .
Right working
$a^{2} = 9$ (denominator of the positive term).
Do this: Use the sign, not the size.
Usually costs: The whole question.
Mistake: Placing foci on the wrong axis
The major axis was decided from the numerator instead of the denominator.
Wrong working
$\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1$ given foci $(\pm 3, 0)$ .
Right working
Larger denominator under $y^{2}$ : foci $(0, \pm 3)$ .
Do this: Compare denominators first.
Usually costs: 1 to 2 marks.
Mistake: Not dividing by the constant to get 1
The equation is not in standard form.
Wrong working
$9 x^{2} + 16 y^{2} = 144$ read as $a^{2} = 9, b^{2} = 16$ .
Right working
Divide by 144: $a^{2} = 16$ , $b^{2} = 9$ .
Do this: Make the right side 1 before reading.
Usually costs: The whole answer.

Exam tips

Put every conic equation into standard form with 1 on the right before reading anything.
Write $a^{2}$ , $b^{2}$ and $c^{2}$ on separate lines with the correct relation for the conic.
Mention the orientation (x-axis or y-axis) explicitly; it earns a method mark.
For equation questions, check your answer with the given point or focus.
Know the difference between the sum (ellipse) and difference (hyperbola) definitions; assertion-reason items test it.

Practice questions on Ellipse and Hyperbola: Foci, Eccentricity and Equations

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 eccentricity of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ is:
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{3}{4}$
(d) $\frac{9}{25}$
Hint
$c^{2} = a^{2} - b^{2}$ .
Answer
(a). $c = 25 - 16 = 3$ , $e = \frac{3}{5}$ .
Q2 · 1 mark · MCQHarder board level
The eccentricity of the hyperbola $x^{2} - y^{2} = 9$ is:
(a) 1
(b) $2$
(c) 2
(d) $\frac{1}{2}$
Hint
Write it as $\frac{x ^{2}}{9} - \frac{y ^{2}}{9} = 1$ .
Answer
(b). $a = b = 3$ , $c = 18 = 32$ , $e = 2$ .
Q3 · 3 marks · Short answerHarder board level
Find the equation of the ellipse with centre at the origin, major axis along the $x$ -axis, passing through $(4, 3)$ and $(6, 2)$ .
Hint
Put $A = a^{2}$ , $B = b^{2}$ and eliminate.
Answer
$\frac{16}{A} + \frac{9}{B} = 1$ , $\frac{36}{A} + \frac{4}{B} = 1$ . Clearing: $16 B + 9 A = A B$ and $36 B + 4 A = A B$ , so $5 A = 20 B$ , $A = 4 B$ . Then $\frac{4}{B} + \frac{9}{B} = 1$ , $B = 13$ , $A = 52$ . Equation $\frac{x ^{2}}{52} + \frac{y ^{2}}{13} = 1$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ has its foci at $(\pm 5, 0)$ .
Reason (R): For a hyperbola $c^{2} = a^{2} + b^{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute $c$ from $a^{2} = 9$ , $b^{2} = 16$ .
Answer
(a). $c^{2} = 9 + 16 = 25$ , so $c = 5$ .
Q5 · 4 marks · Short answerHOTS / competency
The latus rectum of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ ( $a > b$ ) is half of its minor axis. Find the eccentricity. If the latus rectum is 3, write the equation of the ellipse.
Hint
LR $= \frac{2 b ^{2}}{a}$ and half the minor axis is $b$ .
Answer
$\frac{2 b ^{2}}{a} = b$ gives $b = \frac{a}{2}$ . Then $e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{1}{4} = \frac{3}{2}$ . If the LR is $3$ , then $b = 3$ , $a = 6$ , and $\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1$ (check: $c = 27$ , $e = \frac{3 3}{6} = \frac{3}{2}$ ).

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.What is c² for an ellipse?Show answer
a² − b².
2.What is c² for a hyperbola?Show answer
a² + b².
3.What is the eccentricity range of an ellipse?Show answer
0 < e < 1.
4.What is the latus rectum of an ellipse or hyperbola?Show answer
2b²/a.
5.What is the eccentricity of a rectangular hyperbola?Show answer
√2.

Ellipse and Hyperbola: Foci, Eccentricity and Equations: frequently asked questions

How do I tell an ellipse from a hyperbola?

An ellipse has a plus sign between the x² and y² terms; a hyperbola has a minus sign.

Can b be larger than a in an ellipse?

Yes, but then the major axis is along the y-axis and the roles of a and b swap in the formulas for c and e.

What is the hyperbola's transverse axis?

The line segment joining the two vertices; its length is 2a.

Is the circle a special ellipse?

Yes: a = b gives c = 0 and e = 0. The 2026-27 syllabus has removed only the NCERT section on special cases of the ellipse, so the main treatment stays.