Question

The equation of the tangent and normal at point $(3,-2)$ of ellipse $4x^2+9y^2=36$ are

• A. $\frac{x}{3}-\frac{y}{2}=1,\frac{x}{2}+\frac{y}{3}=\frac{5}{6}$
• B. $\frac{x}{3}+\frac{y}{2}=1,\frac{x}{2}-\frac{y}{3}=\frac{5}{6}$
• C. $\frac{x}{2}+\frac{y}{3}=1,\frac{x}{3}-\frac{y}{2}=\frac{5}{6}$
• D. None of the above

Answer 1

The correct option is A $\frac{x}{3}-\frac{y}{2}=1,\frac{x}{2}+\frac{y}{3}=\frac{5}{6}$

Given equation of ellipse is $4x^2+9y^2=36$ or $\frac{x^2}{9}+\frac{y^2}{4}=1$
Tangent at point $(3,-2)$ is $\frac{\left(3\right)x}{9}+\frac{(-2)y}{4}=1$ or $\frac{x}{3}-\frac{y}{2}=1$
$\therefore$ Normal is $\frac{x}{2}+\frac{y}{3}=k$ and it passes through point $(3,-2)$.
Then, $\frac{3}{2}-\frac{2}{3}=k\Rightarrow k=\frac{5}{6}$
Hence, normal is $\frac{x}{2}+\frac{y}{3}=\frac{5}{6}$ and tangent is $\frac{x}{3}-\frac{y}{2}=1$.