JEE Main 2019 — Hyperbola Question with Solution

The equation of a hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its foci are $\left(\pm ae, 0\right)$ and directrices are $x=\pm \frac{a}{e}.$

The given equation of hyperbola is $16x^2-9y^2=144, \Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1$

$\Rightarrow a^2=9, b^2=16$

Given, directrix of the hyperbola is $5x+9=0, \Rightarrow x=-\frac{9}{5}.$

Hence, on comparing the directrix with the standard equation, we get $\frac{a}{e}=\frac{9}{5}$

$\Rightarrow \frac{3}{e}=\frac{9}{5}$

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

And, hence the required focus is $\left(-ae, 0\right)=(-5, 0).$