JEE Mathematics 2025 Question with Solution

Given: The hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. One focus is $\left(\sqrt{10}, 0\right)$ and the corresponding directrix is stated in the solution working as $x = \frac{9}{\sqrt{10}}$.

Find: The value of $9(e^2 + l)$, where $e$ is the eccentricity and $l$ is the length of the latus rectum.

For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,

$$\text{focus} = (\pm ae, 0), \qquad \text{directrix} = x = \pm \frac{a}{e}$$

So,

$$ae = \sqrt{10}$$

and

$$\frac{a}{e} = \frac{9}{\sqrt{10}}$$

Multiplying these two equations,

$$(ae)\left(\frac{a}{e}\right) = \sqrt{10}\cdot \frac{9}{\sqrt{10}}$$ $$a^2 = 9$$

Hence,

$$a = 3$$

Now from $ae = \sqrt{10}$,

$$3e = \sqrt{10}$$ $$e = \frac{\sqrt{10}}{3}$$

Therefore,

$$e^2 = \frac{10}{9}$$

Using the relation

$$b^2 = a^2(e^2 - 1)$$

we get

$$b^2 = 9\left(\frac{10}{9} - 1\right) = 9\left(\frac{1}{9}\right) = 1$$

Now the length of the latus rectum is

$$l = \frac{2b^2}{a} = \frac{2\cdot 1}{3} = \frac{2}{3}$$

Therefore,

$$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right)$$ $$= 9\left(\frac{10}{9} + \frac{6}{9}\right) = 9\left(\frac{16}{9}\right) = 16$$

Hence, the value of $9(e^2 + l)$ is $16$. The correct option is B.

Note: The question statement gives the directrix as $x = \frac{\sqrt{10}}{2}$, but the extracted solution uses $x = \frac{9}{\sqrt{10}}$ and leads consistently to option B.

Given: Focus $= \left(\sqrt{10},0\right)$.

Find: $9(e^2+l)$.

For a horizontal hyperbola,

• focus is $\left(\pm ae,0\right)$
• directrix is $x = \pm \frac{a}{e}$
• $b^2 = a^2(e^2-1)$
• latus rectum length $l = \frac{2b^2}{a}$

From the solution,

$$ae = \sqrt{10}, \qquad \frac{a}{e} = \frac{9}{\sqrt{10}}$$

Multiply:

$$a^2 = 9$$

So,

$$a = 3$$

Then,

$$e = \frac{\sqrt{10}}{3}$$

Thus,

$$e^2 = \frac{10}{9}$$

Now,

$$b^2 = 9\left(\frac{10}{9}-1\right)=1$$

Hence,

$$l = \frac{2b^2}{a} = \frac{2}{3}$$

Finally,

$$9(e^2+l)=9\left(\frac{10}{9}+\frac{2}{3}\right)=9\left(\frac{16}{9}\right)=16$$

Therefore, the correct option is B.