Question

Find the equation of normal to the point on the ellipse
$\frac{x^2}{36}+\frac{y^2}{16}=1$
whose eccentric angle is ${60}^{\circ }$

• A. None of the above
• B. $2x-\sqrt{3}y=5\sqrt{3}$
• C. $3\sqrt{3}x-2y=5\sqrt{3}$
• D. $\sqrt{3}x-2y=\sqrt{3}$

Answer 1

The correct option is C $3\sqrt{3}x-2y=5\sqrt{3}$

$\text{Given ellipse is}:\frac{x^2}{36}+\frac{y^2}{16}=1$
$\text{Here ,}a=6,b=4,\theta ={60}^{\circ }$

Equation of normal in parametric form:
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$
Substituting values of $a,b,\theta$ we get;
$\frac{6x}{\cos {60}^{\circ }}-\frac{4y}{\sin {60}^{\circ }}=36-16$
$\Rightarrow \frac{6x}{\frac{1}{2}}-\frac{4y}{\frac{\sqrt{3}}{2}}=20$
$\Rightarrow 12x-\frac{8y}{\sqrt{3}}=20$
$\Rightarrow 3x-\frac{2y}{\sqrt{3}}=5$
Multiplying both sides by $\sqrt{3},$ we get the equation of normal as:
$3\sqrt{3}x-2y=5\sqrt{3}$
This is the required equation