Question

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

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

Answer 1

Answer: (B)

The equation of the normal at the point $\left(x_1,y_1\right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{x-x_1}{x_1/a^2}=\frac{y-y_1}{y_1/b^2}$
Equation of ellipse is
$9x^2+5y^2=45$
$\frac{x^2}{5}+\frac{y^2}{9}=1$
Here, $a^2=5,b^2=9$
Equation of normal to the ellipse at the point $(0,3)$ is
$\frac{x-0}{0/5}=\frac{y-3}{3/9}$
$\Rightarrow x=0$
Which is the equation of $y$ -axis.
Alternative Given equation is$<br/><br/>9x^2+5y^2=45$
On differentiating, we get
$18x+10y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{18x}{10y}$
At $(0,3),\left(\frac{dy}{dx}\right)=\frac{-18(0)}{10(3)}=0$
$\therefore$ Equation of normal is
$y-3=-\frac{1}{0}(x-0)$
$\Rightarrow x=0$
$\Rightarrow$ y-axis