JEE Mathematics 2025 Question with Solution

Given: The hyperbola is

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

and one focus is at $P(-3,0)$. Hence $c=3$.

Find: The value of $\alpha+\beta$ when

$$a^2b^2=\alpha\sqrt{2}-\beta.$$

For the hyperbola, the foci are $(-c,0)$ and $(c,0)$, so the other focus is $(3,0)$. The latus rectum through this focus has endpoints

$$(3, \tfrac{b^2}{a}) \quad \text{and} \quad (3, -\tfrac{b^2}{a}).$$

Since this latus rectum subtends a right angle at $P(-3,0)$, the angle between the lines joining $P$ to these two endpoints is $90^\circ$.

Using the slope condition for perpendicular lines,

$$\tan 45^\circ = \frac{\tfrac{b^2}{a}}{2c}.$$

With $c=3$, this gives

$$1 = \frac{\tfrac{b^2}{a}}{6}$$

so

$$b^2 = 6a.$$

Also, for the hyperbola,

$$c^2 = a^2 + b^2.$$

Since $c=3$,

$$9 = a^2 + b^2.$$

Substituting $b^2=6a$,

$$9 = a^2 + 6a$$

which gives

$$a^2 + 6a - 9 = 0.$$

Therefore,

$$a = -3 \pm 3\sqrt{2}.$$

Since $a>0$,

$$a = -3 + 3\sqrt{2} = 3(\sqrt{2}-1).$$

Now,

$$a^2b^2 = a^2(6a) = 6a^3.$$

Using $a=3(\sqrt{2}-1)$,

$$a^3 = 27(\sqrt{2}-1)^3 = 27(5\sqrt{2}-7).$$

Hence,

$$a^2b^2 = 6 \cdot 27(5\sqrt{2}-7) = 162(5\sqrt{2}-7) = 810\sqrt{2}-1134.$$

So,

$$\alpha = 810, \qquad \beta = 1134.$$

Therefore,

$$\alpha+\beta = 1944.$$

The final answer is $1944$.

Given: One focus is $P(-3,0)$ and the other focus is $(3,0)$.

Find: The value of $\alpha+\beta$.

The latus rectum through $(3,0)$ is the vertical line through that focus, so its endpoints are

$$A\left(3,\frac{b^2}{a}\right), \qquad B\left(3,-\frac{b^2}{a}\right).$$

We are told that $\angle APB = 90^\circ$.

So the slopes of $PA$ and $PB$ are

$$m_1 = \frac{\frac{b^2}{a}-0}{3-(-3)} = \frac{b^2}{6a},$$ $$m_2 = \frac{-\frac{b^2}{a}-0}{3-(-3)} = -\frac{b^2}{6a}.$$

For perpendicular lines,

$$m_1m_2 = -1.$$

Therefore,

$$\left(\frac{b^2}{6a}\right)\left(-\frac{b^2}{6a}\right) = -1$$

which gives

$$\frac{b^4}{36a^2} = 1.$$

Hence,

$$b^2 = 6a$$

because $a>0$ and $b^2>0$.

Now use

$$c^2 = a^2+b^2 = 9.$$

So,

$$a^2+6a=9$$

that is,

$$a^2+6a-9=0.$$

Thus,

$$a = -3 \pm 3\sqrt{2}.$$

Taking the positive value,

$$a = 3(\sqrt{2}-1).$$

Then

$$a^2b^2 = 6a^3 = 810\sqrt{2}-1134.$$

So $\alpha=810$ and $\beta=1134$, and therefore the required value is $1944$.