JEE Main 2025 — Hyperbola Question with Solution

Given:

Transverse axis length: $2a$

Conjugate axis length: $2b$

One focus at $(-5,0)$

Directrix given by $5x+9=0$

The equations used are as follows:

Relationship between focus and directrix:

The focal length $ae=5$

The directrix gives $\frac{a}{e}=\frac{9}{5}$

Solving these equations, we get:

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

From $ae=5$, we have $a=\frac{5}{e}$.

Substituting $a=3$ and $e=\frac{5}{3}$.

Finding $b$:

Use the relationship $b^2=a^2(e^2-1)$:

$\text{Given }a=3\text{and }e=\frac{5}{3},b=4$

Equation of the hyperbola:

The standard equation, after substituting values of $a$ and $b$, is:

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

Focal distances product calculation:

For any point $(\alpha ,2\sqrt{5})$ that lies on the hyperbola:

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

Solving this gives:

${\alpha }^2=9\times \frac{36}{16}$

Product of focal distances ${\mathrm{PF}}_1\cdot {\mathrm{PF}}_2$:

${\mathrm{PF}}_1\cdot {\mathrm{PF}}_2=(e\alpha -a)(e\alpha +a)=e^2{\alpha }^2-a^2$

Substituting the values:

$P=e^2{\alpha }^2-a^2=\frac{25}{9}\cdot 9\cdot \frac{9}{4}-9=\frac{189}{4}$

Finally, to find $4p$:

$4p=4\times \frac{189}{4}=189$