1. ## Conics Questions Help!!

The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

Thanks

2. Originally Posted by jeta

The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

Thanks
One approach would be to get the equation of the tangent at P. You can do this the same way as your hyperbola question: http://www.mathhelpforum.com/math-he...a-problem.html.

Substitute x = -a into the equation of the tangent to get the y-coord of Q and hence distance AQ.

Substitute x = a into the equation of the tangent to get the y-coord of R and hence distance BR.

Now calculate dAQ.dBR and hope you get b^2 as the answer.

3. Hello,
Originally Posted by jeta

The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

Thanks
I'll let you draw the diagram.

A and B are on the x axis. This means that A and B have their ordinate equal to 0.
---> x²/a² + 0=1 ---> $x=\pm a$

So let $\boxed{A(a,0)}$ and $\boxed{B(-a,0)}$ (the coordinates can be inverted, but I choose them this way. It has no importance)

If you do an implicit differentiation of the ellipse's equation and that you substitute y=0, you will see that y'=infinity --> vertical

An equation of a vertical line is in the form x=constant.
So here, the vertical tangent to the ellipse at A is $\boxed{x=a}$, since A is on this tangent. Similarly, the vertical tangent to the ellipse at B is $\boxed{x=-a}$.

The equation of the tangent at point P is :
$y=y'(a \cos(t))(x-(a \cos(t)))+\underbrace{y(a \cos(t))}_{a \sin(t)}$

By implicit differentiation and rearranging, we have $y'(t)=\frac{-b^2 x'(t) x(t)}{a^2 y(t)}$
We know that at point P, $x(t)=a \cos(t)$ and $y(t)=b \sin(t)$. And $x'(t)=-a \sin(t)$

Thus $y'(a \cos(t))=\frac{b^2a^2 \sin(t) \cos(t)}{a^3 \sin(t)}=\frac{b^2}{a}$

The equation of the tangent at point P is now :
$y=\frac{b^2}{a}(x-a \cos(t))+a \sin(t)=\boxed{\frac{b^2}{a} x-b^2 \cos(t)+a \sin(t)}$

Q is the intersection of this tangent with x=a.
So $y_Q=b^2-b^2 \cos(t)+a \sin(t)$
$\boxed{Q(a,b^2-b^2 \cos(t)+a \sin(t))}$

Similarly, $x_R=-a$ and $y_R=-b^2-b^2 \cos(t)+a \sin(t)$
$\boxed{R(-a,-b^2-b^2 \cos(t)+a \sin(t))}$

Now, find $AQ=\sqrt{(x_A-x_Q)^2+(y_A-y_Q)^2}$ and $BR=\dots$

Edit : maybe there's a shorter way, but I decided to do it the straightforward way lol

4. Thanks alot.