Mid-point of chord on ellipse

Quote:

Originally Posted by arze

Prove that the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $P_1(x_1,y_1)$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ . The tangent at $P_1$ meets the tangent at $P_2(x_2,y_2)$ at T. Show that the line
$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$
passes through T and through the midpoint of $P_1P_2$ . Prove that if $P_1TP_2$ is a right angle, then $\frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0$

I have proven the first part. I got the equation of the tangent at $P_2$ in the same form as the first. Then I made y the subject of both tangents and found x. did it in the same way to find y.
I found the the line given passed through T, but could not prove that the line passed through the midpoint of $P_1P_2$ , $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ .

The equation $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$ represents a straight line. If the point (x,y) lies on the tangent at $P_1$ then the left side of that equation will be 1. If (x,y) lies on the tangent at $P_2$ then the right side of the equation will be 1. If (x,y) is the point T then both sides of the equation will be 1, and hence the equation will be satified. Therefore T lies on the line with that equation.

The condition for the midpoint of $P_1P_2$ to lie on the line is $\frac{\frac12(x_1+x_2)x_1}{a^2}+\frac{\frac12(y_1+ y_2)y_1}{b^2}=\frac{\frac12(x_1+x_2)x_2}{a^2}+\fra c{\frac12(y_1+y_2)y_2}{b^2}$ . After multiplying out all the brackets and doing some cancellation, you should be able to write that condition as $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} = \frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}$ . Both sides of that equation are equal to 1 because $P_1$ and $P_2$ lie on the ellipse. Therefore the condition is satisfied, and the midpoint of $P_1P_2$ lies on the line.

For the last part of the question, write down the condition for the tangents at $P_1$ and $P_2$ to have perpendicular slopes.