# Ellipse

• February 23rd 2010, 01:06 AM
Sunyata
Ellipse
Hi

How exactly would I do this?

The normal to the ellipse x^2/a^2 + y^2/b^2 = 1 at P(x1, y1) meets the x-axis in N and the Y-axis in G. Prove PN/NG = (1-e^2)/e^2
• February 23rd 2010, 01:20 AM
Opalg
See this thread for some general comments about ellipses that should help with this problem. Remember that the eccentricity is given by $e^2 = 1 - \frac{b^2}{a^2}$.
• February 26th 2010, 04:38 PM
Sunyata
Hi

• February 27th 2010, 07:56 AM
Opalg
Quote:

Originally Posted by Sunyata
I've spent a few days on this but I cant solve it.

I'm not entirely surprised. This is a longer and messier calculation than I would have guessed. Here is an outline of how to do it.

Take P to be the point $(a\cos\theta,b\sin\theta)$. Then (see the link in my other comment above) the normal at P has equation $yb\cos\theta - xa\sin\theta = (b^2 - a^2)\cos\theta\sin\theta$.

Put y = 0 in that equation to see that N is the point $\Bigl(\frac{(a^2-b^2)\cos\theta}a,0\Bigr)$. Put x = 0 to see that G is the point $\Bigl(0,\frac{(b^2-a^2)\sin\theta}b\Bigr)$.

Then use the usual distance formula to see that $PN^2 = \Bigl(\frac{b^2\cos\theta}a\Bigr)^2 + b^2\sin^2\theta = \frac{b^2(b^2\cos^2\theta + a^2\sin^2\theta)}{a^2}$.

Similarly $NG^2 = (a^2-b^2)^2\Bigl(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\Bigr) = \frac{(a^2-b^2)^2(b^2\cos^2\theta + a^2\sin^2\theta)}{a^2b^2}$.

Therefore $\Bigl(\frac{PN}{NG}\Bigr)^2 = \frac{b^4}{(a^2-b^2)^2}$, and so $\frac{PN}{NG} = \frac{b^2}{a^2-b^2} = \frac{a^2(1-e^2)}{a^2e^2} = \frac{1-e^2}{e^2}$.