Question

The equation of the normal at the positive end of the latusrectum of the hyperbola $x^2-3y^2=144$ is

• A. $\sqrt{3}x+2y=32$
• B. $\sqrt{3}x-3y=48$
• C. $3x+\sqrt{3}y=48$
• D. $3x-\sqrt{3}y=48$

Answer 1

Answer: (A)

$\sqrt{3}x+2y=32$

The given hyperbola has the equation $\frac{x^2}{{12}^2}-\frac{y^2}{(4\sqrt{3}{)}^2}=1$

Eccentricity of hyperbola = $e=\frac{\sqrt{a^2+b^2}}{a}=\frac{2}{\sqrt{3}}$

Now, equation of positive latus rectum is $x=ae=8\sqrt{3}$

The end-points of latus rectum are calculated as

$\frac{(8\sqrt{3}{)}^2}{{12}^2}-\frac{y^2}{(4\sqrt{3}{)}^2}=1$

$\therefore \frac{16}{12}-\frac{y^2}{48}=1$

$\therefore y^2=\frac{48}{3}$

$\therefore y=\pm 4$

Hence, the positive end is $(8\sqrt{3},4)$.

Now, equation of normal at any point $(x_1,y_1)$ is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2{e}^2$

$\therefore \frac{144x}{8\sqrt{3}}+\frac{48y}{4}=48\times 4$

$\therefore 6\sqrt{3}x+12y=48\times 4$

$\therefore \sqrt{3}x+2y=32$

This is the required answer.