Conics problem....need help (1 Viewer)

[linxicm]

P and Q are the point on the ellipse x2/a2 + y2/b2 = 1 and the chord PQ contains a foucs. show that the line joining the point of intersection of the normals at P and Q and the midpont of the chord PQ is parallar to the x axis.

Can someone help me to solve this problem? thanks........

 

[withoutaface]

Ok for this I cbb deriving the normal equations, because you can do that yourself fairly easily.
P=(x₁,y₁), Q=(x₂, y₂)
The equation to the normal at P is:
a²x/x₁-b²y/y₁=b²-a²

And the midpoint of the chord of the ellipse can be found thus:

((x₁+x₂)/2, (y₁+y₂)/2)

Now all we need to prove is that the point of intersection has the y coordinate (y₁+y₂)/2.

You then sub that into the normal equations for both P and Q and prove that this gives the same x-coordinate for both.

EDIT: The substitution will probably get abit messy, but it should come out.

 

[haboozin]

this means that X1 and X2 are both ae correct?

i did it assuming that...
and my y intercept is 0......
oh i just realized the midpoint of the 2 points should be the focus.. haha oops
so yea it should be 0..



therefore there is no rise... and the gradient is 0 which is the same as the x-axis