Coordinate Geometry – Hyperbola – JEE Main 2 Apr 2025 Shift 1

Solution:

For the standard hyperbola $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$:
$$ae = \sqrt{10} \quad \text{and} \quad \frac{a}{e} = \frac{9}{\sqrt{10}}$$

Divide the two equations to eliminate $a$:
$$\frac{ae}{a/e} = \frac{\sqrt{10}}{9/\sqrt{10}} \Rightarrow e^2 = \frac{10}{9} \Rightarrow e = \frac{\sqrt{10}}{3}$$

Substitute $e$ back to find $a$:
$$a = \frac{\sqrt{10}}{e} = \frac{\sqrt{10}}{\sqrt{10}/3} = 3 \Rightarrow a^2 = 9$$

Use the relation $b^2 = a^2(e^2 – 1)$:
$$b^2 = 9\left(\frac{10}{9} – 1\right) = 9\left(\frac{1}{9}\right) = 1$$

The length of the latus rectum $l$ is given by:
$$l = \frac{2b^2}{a} = \frac{2(1)}{3} = \frac{2}{3}$$

Now, calculate the required value:
$$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right)$$

$$= 10 + 6 = 16$$

Ans. (3)