Normal of a point on an ellipse

[Appleton]

Homework Statement

11) The tangent at P on the ellipse $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$ 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

$y=\frac {a\sin\theta}{b\cos\theta}x-\frac{(a^2-b^2)\sin\theta}{b}$

When y = 0

$x=\frac{(a^2-b^2)\cos\theta}{a}$

So

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

When x = 0

$y=-\frac{(a^2-b^2)\sin\theta}{b}$

So

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

$AP^2= (a\cos\theta-\frac{(a^2-b^2)\cos\theta}{ a})^2+ (b\sin\theta)^2$

$BP^2= (b\sin\theta+\frac{(a^2-b^2)\sin\theta}{ b})^2 + (a\cos\theta)^2$

$\frac{AP^2}{BP^2}=\frac{b^4(a^2\sin^2\theta+b^2\cos^2\theta)}{ a^4(a^2\sin^2\theta+b^2\cos^2\theta)}$

$\frac{AP}{BP}=\frac{b^2}{a^2}$ = the ratio of the lengths AP to BP

However my textbook says the correct answer is
$\frac{a^2}{b^2}$

 

[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:

$\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$

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?