JEE Mathematics 2025 Question with Solution

Given: The hyperbola is in standard form with one focus at $(-5, 0)$ and the corresponding directrix $5x + 9 = 0$, that is, $x = -\frac{9}{5}$. A point $(\alpha, 2\sqrt{5})$ lies on the hyperbola.

Find: The value of $4p$, where $p$ is the product of the focal distances of the point from the two foci.

For the standard hyperbola

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

the foci are $(\pm ae, 0)$ and the directrices are $x = \pm \frac{a}{e}$.

Since one focus is $(-5, 0)$, we have

$$ae = 5$$

And since the corresponding directrix is $x = -\frac{9}{5}$, we get

$$\frac{a}{e} = \frac{9}{5}$$

Multiplying,

$$a^2 = (ae)\left(\frac{a}{e}\right) = 5 \times \frac{9}{5} = 9$$

So,

$$a = 3, \qquad e = \frac{5}{3}$$

Now use

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

Substituting the values,

$$\left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{9}$$ $$\frac{25}{9} - 1 = \frac{b^2}{9}$$ $$\frac{16}{9} = \frac{b^2}{9}$$ $$b^2 = 16$$

Hence the hyperbola is

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

Since $(\alpha, 2\sqrt{5})$ lies on it,

$$\frac{\alpha^2}{9} - \frac{(2\sqrt{5})^2}{16} = 1$$ $$\frac{\alpha^2}{9} - \frac{20}{16} = 1$$ $$\frac{\alpha^2}{9} = \frac{9}{4}$$ $$\alpha^2 = \frac{81}{4}$$

Thus,

$$\alpha = \pm \frac{9}{2}$$

The foci are $(\pm 5, 0)$. So the focal distances are

$$d_1 = \sqrt{(\alpha + 5)^2 + (2\sqrt{5})^2}, \qquad d_2 = \sqrt{(\alpha - 5)^2 + (2\sqrt{5})^2}$$

Taking $\alpha = \frac{9}{2}$,

$$d_1 = \sqrt{\left(\frac{9}{2} + 5\right)^2 + 20} = \sqrt{\left(\frac{19}{2}\right)^2 + 20} = \sqrt{\frac{361}{4} + \frac{80}{4}} = \sqrt{\frac{441}{4}} = \frac{21}{2}$$ $$d_2 = \sqrt{\left(\frac{9}{2} - 5\right)^2 + 20} = \sqrt{\left(-\frac{1}{2}\right)^2 + 20} = \sqrt{\frac{1}{4} + \frac{80}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$$

Therefore,

$$p = d_1 d_2 = \frac{21}{2} \times \frac{9}{2} = \frac{189}{4}$$

Hence,

$$4p = 189$$

Therefore, the required value is $189$.

Given: Focus $(-5, 0)$ and corresponding directrix $x = -\frac{9}{5}$.

Find: $4p$.

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

$$ae = 5, \qquad \frac{a}{e} = \frac{9}{5}$$

So,

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

which gives $a = 3$. Then

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

and hence

$$b^2 = a^2(e^2 - 1) = 9\left(\frac{25}{9} - 1\right) = 16$$

So the hyperbola is

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

Putting $y = 2\sqrt{5}$,

$$\frac{x^2}{9} - \frac{20}{16} = 1$$ $$\frac{x^2}{9} = \frac{9}{4}$$ $$x = \pm \frac{9}{2}$$

Now the distances to the foci are $\frac{21}{2}$ and $\frac{9}{2}$, so

$$p = \frac{21}{2} \cdot \frac{9}{2} = \frac{189}{4}$$

Therefore,

$$4p = 189$$

The required value is $189$.