Question

The equation of the normal to the ellipse $\frac{x^2}{18}+\frac{y^2}{8}=1$ at the point (3,2) is .

• A. 3x - 2y = 5
• B. 3x + 2y = 5
• C. 2x - 3y = 5
• D. 2x + 3y = 5

Answer 1

The correct option is A 3x - 2y = 5

$\mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{of}\mathrm{ellipse}\mathrm{is}$ $\frac{x^2}{18}+\frac{y^2}{8}=1$.
$\mathrm{Differentiating}w.r.tx,\mathrm{we}\mathrm{get}:$
$\frac{2x}{18}+\frac{2y}{8}\frac{\mathrm{dy}}{\mathrm{dx}}=1\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4x}{9y}\therefore \mathrm{Slope}\mathrm{of}a\mathrm{normal}=-\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{9y}{4x}\therefore \mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{at}\mathrm{the}\mathrm{point}(3,2)=\frac{9\left(2\right)}{4\left(3\right)}=\frac{3}{2}\therefore \mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{will}\mathrm{be}:y-2=\frac{3}{2}(x-3)\mathrm{or}3x-2y=5$