Normal of a point on an ellipse

[Appleton]

Homework Statement

11) The tangent at P on the ellipse \frac{x^2}{a^2}+<br /> \frac{y^2}{b^2}=1 meets the x and y axes at A and B.
Find, in terms of the eccentric angle of P, the ratio of the lengths AP and BP.

12) Repeat Question 11 using the normal at P.

Homework Equations

bx \cos \theta + ay \sin \theta = ab

The Attempt at a Solution

[/B]
I had no significant difficulty with 11, my problem was with 12:

The normal at P is

y=\frac {a\sin\theta}{b\cos\theta}x-\frac{(a^2-b^2)\sin\theta}{b}<br />

When y = 0

x=\frac{(a^2-b^2)\cos\theta}{a}<br />

So

A =\left(\begin{array}{cc}\frac{(a^2-b^2)\cos\theta}{a}&amp;0\end{array}\right)

When x = 0

y=-\frac{(a^2-b^2)\sin\theta}{b}<br />

So

B =\left(\begin{array}{cc}0&amp;-\frac{(a^2-b^2)\sin\theta}{b}\end{array}\right)

AP^2= (a\cos\theta-\frac{(a^2-b^2)\cos\theta}{<br /> a})^2+<br /> (b\sin\theta)^2<br />

BP^2= (b\sin\theta+\frac{(a^2-b^2)\sin\theta}{<br /> b})^2 +<br /> (a\cos\theta)^2<br />

\frac{AP^2}{BP^2}=\frac{b^4(a^2\sin^2\theta+b^2\cos^2\theta)}{<br /> a^4(a^2\sin^2\theta+b^2\cos^2\theta)}<br />

\frac{AP}{BP}=\frac{b^2}{a^2}<br /> <br /> = the ratio of the lengths AP to BP

However my textbook says the correct answer is
\frac{a^2}{b^2}<br /> <br />

 

[jedishrfu]

Check to see if you've used a and b the same way as your textbook.

I noticed in your response to 2. that you have $bx cos \theta + ay sin \theta = ab$

Perhaps your book has defined it as ax and by instead

 

[Appleton]

Hi jedishrfu, thanks for your reply, I hadn't considered that. However the equation of the tangent does seem to be quoted from the book correctly, although I used the following rearranged one, also quoted in the book, in my calculations:

<br /> \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1<br />

If i have quoted the equation of the tangent from the book correctly can I put this error down to a typo in the answers section of the book?