Question

Find the parametric equation of the normal to the ellipse: \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] at the point \[P(a\cos(\theta), b\sin(\theta))\]

Answer 1

I got an answer that's correct. But I need a more simplified equation if possible.

Answer 2

Whats your progress ?

Answer 3

Implicit differentiation.
\[D_{x}(\frac{x^2}{a^2}+\frac{y^2}{b^2})=0\]
\[\frac{2x}{a^2}+\frac{2y}{b^2}(\frac{dy}{dx})=0\]
\[\frac{dy}{dx}=-\frac{2x}{a^2}\times \frac{b^2}{2y}\]
\[=-\frac{b^2x}{a^2y}\]
At \[P(a\cos\theta, b\sin\theta)\]
\[\large m_{T}=-\frac{b^2a\cos\theta}{a^2b\sin\theta}\]
\[=-\frac{b\cos\theta}{a\sin\theta}\]
Therefore
\[\large m_{N}=\frac{a\sin\theta}{b\cos\theta}\]
To be continued...

Answer 4

\[y-b\sin\theta=\frac{a\sin\theta}{b\cos\theta}(x-a\cos\theta)\]
\[by\cos\theta -b^2\sin\theta\cos\theta=ax\sin\theta-a^2\cos\theta\sin\theta\]
\[ax\sin\theta-by\cos\theta=a^2\sin\theta\cos\theta-b^2\sin\theta\cos\theta\]
\[=\sin\theta\cos\theta(a^2-b^2)\]
I was going to substitute in b^2 with this equation \[b^2=a^2(1-e^2)\] But then I would just bring the eccentricity into the equation which I don't want. But maybe that would work. Not sure.

Answer 5

@shubhamsrg

Answer 6

Well why do you want to simplify it ? You have done absolutely fine till now.

Answer 7

Not simply, modify *