JEE Mathematics 2025 Question with Solution

Given: The hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the point is $P(4,3)$.

Find: The value of $9\ell^2 + 6m$, where $\ell$ is the length of the latus rectum and $m$ is the product of the focal distances of $P$.

From the solution, the sum of the focal distances is used as

$$2ex = 8\sqrt{\frac{5}{3}}$$

At $x=4$,

$$2e \cdot 4 = 8\sqrt{\frac{5}{3}}$$

so

$$e = \sqrt{\frac{5}{3}}$$

Hence,

$$e^2 = \frac{5}{3}$$

Using the relation for hyperbola,

$$e^2 = 1 + \frac{b^2}{a^2}$$

therefore

$$\frac{b^2}{a^2} = \frac{5}{3} - 1 = \frac{2}{3}$$

so

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

Now the point $P(4,3)$ lies on the hyperbola, hence

$$\frac{16}{a^2} - \frac{9}{b^2} = 1$$

Substituting $b^2 = \frac{2a^2}{3}$,

$$\frac{16}{a^2} - \frac{9}{\frac{2a^2}{3}} = 1$$ $$\frac{16}{a^2} - \frac{27}{2a^2} = 1$$ $$\frac{32-27}{2a^2} = 1$$ $$\frac{5}{2a^2} = 1$$

Therefore,

$$a^2 = \frac{5}{2}$$

and then

$$b^2 = \frac{2}{3} \cdot \frac{5}{2} = \frac{5}{3}$$

The length of the latus rectum is

$$\ell = \frac{2b^2}{a}$$

So,

$$\ell^2 = \frac{4b^4}{a^2}$$

Substituting $a^2 = \frac{5}{2}$ and $b^2 = \frac{5}{3}$,

$$\ell^2 = \frac{4\left(\frac{5}{3}\right)^2}{\frac{5}{2}} = \frac{40}{9}$$

Hence,

$$9\ell^2 = 40$$

For the product of focal distances, the solution uses

$$m = (ex+a)(ex-a)$$

At $x=4$,

$$m = (4e+a)(4e-a) = 16e^2 - a^2$$

Substituting $e^2 = \frac{5}{3}$ and $a^2 = \frac{5}{2}$,

$$m = 16\cdot \frac{5}{3} - \frac{5}{2} = \frac{145}{6}$$

Therefore,

$$9\ell^2 + 6m = 40 + 145 = 185$$

So, the correct option is C.

Given: $P(4,3)$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the sum of focal distances is $8\sqrt{\frac{5}{3}}$.

Find: $9\ell^2 + 6m$.

For a point $P(x,y)$ on the hyperbola, the focal distances are taken in the solution as

$$ex+a \quad \text{and} \quad ex-a$$

So their sum is

$$(ex+a) + (ex-a) = 2ex$$

Given this sum is $8\sqrt{\frac{5}{3}}$, we get

$$2ex = 8\sqrt{\frac{5}{3}}$$

Now put $x=4$:

$$8e = 8\sqrt{\frac{5}{3}}$$ $$e = \sqrt{\frac{5}{3}}$$

Thus,

$$e^2 = \frac{5}{3}$$

Also,

$$e^2 = 1 + \frac{b^2}{a^2}$$

So,

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

Since $P(4,3)$ lies on the hyperbola,

$$\frac{16}{a^2} - \frac{9}{b^2} = 1$$

Substitute $b^2 = \frac{2a^2}{3}$:

$$\frac{16}{a^2} - \frac{9}{\frac{2a^2}{3}} = 1$$ $$\frac{16}{a^2} - \frac{27}{2a^2} = 1$$ $$\frac{32}{2a^2} - \frac{27}{2a^2} = 1$$ $$\frac{5}{2a^2} = 1$$ $$a^2 = \frac{5}{2}$$

Then,

$$b^2 = \frac{2}{3}\cdot \frac{5}{2} = \frac{5}{3}$$

Now compute the latus rectum:

$$\ell = \frac{2b^2}{a}$$

Therefore,

$$\ell^2 = \frac{4b^4}{a^2} = \frac{4\left(\frac{5}{3}\right)^2}{\frac{5}{2}} = \frac{40}{9}$$

Hence,

$$9\ell^2 = 40$$

Next,

$$m = (ex+a)(ex-a)$$

Using $x=4$,

$$m = 16e^2 - a^2$$

Substituting values,

$$m = 16\cdot \frac{5}{3} - \frac{5}{2} = \frac{160-15}{6} = \frac{145}{6}$$

Thus,

$$6m = 145$$

Finally,

$$9\ell^2 + 6m = 40 + 145 = 185$$

Therefore, the value is $185$ and the correct option is C.