Question

The equation of the normal to the curve $\frac{x^2}{16}-\frac{y^2}{9}=1$ at $(8,3\sqrt{3})$ is

• A. $4x-\sqrt{3}y=23$
• B. $2x+\sqrt{3}y=25$
• C. $3x-2\sqrt{3}y=6$
• D. $x+\sqrt{3}y=17$

Answer 1

The correct option is B $2x+\sqrt{3}y=25$

Equation of normal in point form is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$
$\therefore$ equation of the normal to the curve $\frac{x^2}{16}-\frac{y^2}{9}=1$ at $(8,3\sqrt{3})$ is
$\frac{16x}{8}+\frac{9y}{3\sqrt{3}}=16+9$
$\Rightarrow 2x+\sqrt{3}y=25$
Hence, the required equation of the normal is $2x+\sqrt{3}y=25.$