Question

What is the equation of the normal with slope m to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1?$

• A. $y=mx-\frac{(a^2+b^2)m}{\sqrt{a^2+b^2{m}^2}}$
• B. $y=mx-\frac{(a^2+b^2)m}{\sqrt{a^2+m^2+b^2}}$
• C. $y=mx-\frac{(a^2-b^2)m}{\sqrt{a^2+b^2{m}^2}}$
• D. $y=mx-\frac{m(a^2+b^2)}{\sqrt{a^2{m}^2+b^2}}$

Answer 1

The correct option is C

$y=mx-\frac{(a^2-b^2)m}{\sqrt{a^2+b^2{m}^2}}$

This is a result which will save a lot of time if you can remember it. We will derive it in this solution.
Any line with slope m can be written as y = mx + c. ...(i)
We want to find the value of C.
The equation of normal at $(acos\theta ,bsin\theta )$ is
$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=a^2-b^2$
We can rearrange it as
Comparing (1) and (2) we get
$\frac{a}{b}tan\theta =mand-(a^2-b^2)\frac{sin\theta }{b}=c\Rightarrow Tofindc,wehavetofindsin\theta Tan\theta =\frac{bm}{a}$

$\Rightarrow sin\theta =\frac{bm}{\sqrt{a^2+b^2{m}^2}}\Rightarrow c=\frac{-(a^2-b^2)}{b}x\frac{bm}{\sqrt{a^2+b^2{m}^2}}\Rightarrow y=mx-\frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}$ is the equation of normal.