Question

The equation of normal at the point $(0,3)$ of the ellipse $9x^2+5y^2=45$ is

• A. $y-3=0$
• B. $y+3=0$
• C. $x$-axis
• D. $y$-axis

Answer 1

The correct option is C $x$-axis

$9x^2+5y^2=45$

Differentiating with respect to $x$, we get,

$18x+10y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\frac{-0}{30}=0$

Slope of normal = $\frac{1}{0}$

The equation of normal is :

$y-3=\frac{1}{0}(x-0)$

$x-0=0$

$x=0$