NCERT Class 11 Mathematics — Conic Sections, Exercise 10.4. All 15 questions solved.

A hyperbola is the set of points whose distances from two fixed points (the foci) have a constant difference. Two standard forms, centred at the origin:

Form	Transverse axis	Foci	Vertices
$\dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1$	along $x$	$(\pm c, 0)$	$(\pm a, 0)$
$\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$	along $y$	$(0, \pm c)$	$(0, \pm a)$

with

$$c^2 = a^2 + b^2, \qquad e = \frac ca > 1, \qquad \text{latus rectum} = \frac{2b^2}{a}$$

The transverse axis has length $2a$ and the conjugate axis $2b$.

Key insight. For a hyperbola it is the positive term that names the transverse axis — not the larger denominator, as it was for the ellipse. In question 2, $\tfrac{y^2}{9} – \tfrac{x^2}{27} = 1$ has $27 > 9$, but the positive term is the $y$ one, so $a^2 = 9$ and the transverse axis is along $y$. Carrying the ellipse’s “bigger denominator wins” rule across is the single most common error in this exercise.

In questions 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

Question 1

$\dfrac{x^2}{16} – \dfrac{y^2}{9} = 1$

Solution. The $x$ term is positive, so $a^2 = 16$, $b^2 = 9$ and the transverse axis is along the $x$-axis.

$$a = 4, \qquad b = 3, \qquad c = \sqrt{16 + 9} = 5$$

$$e = \frac54, \qquad \text{latus rectum} = \frac{2(9)}{4} = \frac92$$

Question 2

$\dfrac{y^2}{9} – \dfrac{x^2}{27} = 1$

Solution. The positive term is the $y$ one, so $a^2 = 9$ and $b^2 = 27$ — even though $27$ is larger. Transverse axis along $y$.

$$a = 3, \qquad b = 3\sqrt3, \qquad c = \sqrt{9 + 27} = 6$$

$$e = \frac63 = 2, \qquad \text{latus rectum} = \frac{2(27)}{3} = 18$$

Question 3

$9y^2 – 4x^2 = 36$

Solution. Divide by $36$ to reach standard form:

$$\frac{y^2}{4} – \frac{x^2}{9} = 1$$

$a^2 = 4$, $b^2 = 9$, transverse axis along $y$:

$$a = 2, \qquad c = \sqrt{4 + 9} = \sqrt{13}$$

$$e = \frac{\sqrt{13}}{2}, \qquad \text{latus rectum} = \frac{2(9)}{2} = 9$$

Question 4

$16x^2 – 9y^2 = 576$

Solution. Dividing by $576$:

$$\frac{x^2}{36} – \frac{y^2}{64} = 1$$

$a = 6$, $b = 8$, $c = \sqrt{36 + 64} = 10$; transverse axis along $x$.

$$e = \frac{10}{6} = \frac53, \qquad \text{latus rectum} = \frac{2(64)}{6} = \frac{64}{3}$$

Question 5

$5y^2 – 9x^2 = 36$

Solution. Dividing by $36$:

$$\frac{y^2}{36/5} – \frac{x^2}{4} = 1$$

$a^2 = \tfrac{36}{5}$ and $b^2 = 4$, transverse axis along $y$:

$$a = \frac{6}{\sqrt5}, \qquad c = \sqrt{\frac{36}{5} + 4} = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt5}$$

$$e = \frac ca = \frac{2\sqrt{14}/\sqrt5}{6/\sqrt5} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}$$

$$\text{latus rectum} = \frac{2(4)}{6/\sqrt5} = \frac{8\sqrt5}{6} = \frac{4\sqrt5}{3}$$

Question 6

$49y^2 – 16x^2 = 784$

Solution. Dividing by $784$:

$$\frac{y^2}{16} – \frac{x^2}{49} = 1$$

$a = 4$, $b = 7$, $c = \sqrt{16 + 49} = \sqrt{65}$; transverse axis along $y$.

$$e = \frac{\sqrt{65}}{4}, \qquad \text{latus rectum} = \frac{2(49)}{4} = \frac{49}{2}$$

In questions 7 to 15, find the equation of the hyperbola satisfying the given conditions.

Question 7

Vertices $(\pm2, 0)$, foci $(\pm3, 0)$.

Solution. Both on the $x$-axis, so $a = 2$, $c = 3$ and

$$b^2 = c^2 – a^2 = 9 – 4 = 5$$

$$\frac{x^2}{4} – \frac{y^2}{5} = 1$$

Note the rearrangement: for a hyperbola $b^2 = c^2 – a^2$, whereas for an ellipse it was $b^2 = a^2 – c^2$.

Question 8

Vertices $(0, \pm5)$, foci $(0, \pm8)$.

Solution. Transverse axis along $y$: $a = 5$, $c = 8$, $b^2 = 64 – 25 = 39$.

$$\frac{y^2}{25} – \frac{x^2}{39} = 1$$

Question 9

Vertices $(0, \pm3)$, foci $(0, \pm5)$.

Solution. $a = 3$, $c = 5$, $b^2 = 25 – 9 = 16$.

$$\frac{y^2}{9} – \frac{x^2}{16} = 1$$

Question 10

Foci $(\pm5, 0)$, the transverse axis is of length $8$.

Solution. Transverse axis $= 2a = 8$, so $a = 4$; and $c = 5$:

$$b^2 = 25 – 16 = 9$$

$$\frac{x^2}{16} – \frac{y^2}{9} = 1$$

Question 11

Foci $(0, \pm13)$, the conjugate axis is of length $24$.

Solution. The conjugate axis is $2b$, so $b = 12$; and $c = 13$:

$$a^2 = c^2 – b^2 = 169 – 144 = 25$$

The foci on the $y$-axis put $a^2$ under $y^2$:

$$\frac{y^2}{25} – \frac{x^2}{144} = 1$$

Question 12

Foci $\left(\pm3\sqrt5, 0\right)$, the latus rectum is of length $8$.

Solution. From the foci, $c^2 = 45$, and the latus rectum gives

$$\frac{2b^2}{a} = 8 \quad\Longrightarrow\quad b^2 = 4a$$

Substituting into $c^2 = a^2 + b^2$:

$$45 = a^2 + 4a \quad\Longrightarrow\quad a^2 + 4a – 45 = 0 \quad\Longrightarrow\quad (a+9)(a-5) = 0$$

Since $a > 0$, take $a = 5$, and then $b^2 = 20$.

$$\frac{x^2}{25} – \frac{y^2}{20} = 1$$

Question 13

Foci $(\pm4, 0)$, the latus rectum is of length $12$.

Solution. $c = 4$, and

$$\frac{2b^2}{a} = 12 \quad\Longrightarrow\quad b^2 = 6a$$

$$16 = a^2 + 6a \quad\Longrightarrow\quad a^2 + 6a – 16 = 0 \quad\Longrightarrow\quad (a+8)(a-2) = 0$$

So $a = 2$ and $b^2 = 12$.

$$\frac{x^2}{4} – \frac{y^2}{12} = 1$$

Question 14

Vertices $(\pm7, 0)$, $e = \dfrac43$.

Solution. $a = 7$, and from $e = \dfrac ca$:

$$c = ae = 7 \times \frac43 = \frac{28}{3}$$

$$b^2 = c^2 – a^2 = \frac{784}{9} – 49 = \frac{784 – 441}{9} = \frac{343}{9}$$

$$\frac{x^2}{49} – \frac{y^2}{343/9} = 1 \quad\Longrightarrow\quad \frac{x^2}{49} – \frac{9y^2}{343} = 1$$

Question 15

Foci $\left(0, \pm\sqrt{10}\right)$, passing through $(2, 3)$.

Solution. The foci are on the $y$-axis, so the form is

$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1, \qquad c^2 = a^2 + b^2 = 10$$

Substituting the point $(2,3)$, and writing $b^2 = 10 – a^2$:

$$\frac{9}{a^2} – \frac{4}{10 – a^2} = 1$$

Multiply through by $a^2(10 – a^2)$:

$$9(10 – a^2) – 4a^2 = a^2(10 – a^2)$$

$$90 – 13a^2 = 10a^2 – a^4 \quad\Longrightarrow\quad a^4 – 23a^2 + 90 = 0$$

Factorising as a quadratic in $a^2$:

$$\left(a^2 – 5\right)\left(a^2 – 18\right) = 0$$

The root $a^2 = 18$ must be rejected, since $a^2 < c^2 = 10$ for a hyperbola. So $a^2 = 5$ and $b^2 = 10 – 5 = 5$.

$$\frac{y^2}{5} – \frac{x^2}{5} = 1$$

(This is a rectangular hyperbola, with $a = b$.)

Common mistakes

• Taking the larger denominator as $a^2$. For a hyperbola the positive term identifies $a^2$, whatever its size. Question 2, with $9$ under $y^2$ and $27$ under $x^2$, is set precisely to test this.
• Using $c^2 = a^2 – b^2$. That is the ellipse relation. Hyperbolas satisfy $c^2 = a^2 + b^2$, so $c > a$ and $e > 1$ — a useful check on every answer.
• Questions 3 to 6, not dividing by the constant first. $9y^2 – 4x^2 = 36$ must become $\tfrac{y^2}{4} – \tfrac{x^2}{9} = 1$ before anything can be read off.
• Question 11, confusing the transverse and conjugate axes. The transverse axis is $2a$ and the conjugate is $2b$. Here the conjugate is given, so $b = 12$, and $a$ has to be found.
• Questions 12 and 13, keeping the negative root. Both quadratics have one negative solution for $a$, which is geometrically meaningless.
• Question 15, keeping $a^2 = 18$. Since $c^2 = 10$ and $c^2 = a^2 + b^2$ with $b^2 > 0$, $a^2$ must be less than $10$. Checking the root against the geometry is what discards it.
• Question 5, mishandling the fractional $a^2$. $\tfrac{36}{5}$ is a perfectly ordinary $a^2$; $a = \tfrac{6}{\sqrt5}$, and the eccentricity simplifies neatly because the $\sqrt5$ cancels.

Practise next

• Miscellaneous Exercise on Chapter 10 — applied conic problems: parabolic reflectors, semi-elliptical arches and suspension bridges.
• Exercise 10.3 — worth working alongside this one, since the two sets of formulae differ only in the sign of $b^2$ and are easiest to learn by contrast.