conics (1 Viewer)

[gronkboyslim]

P(a cos@, bsin @) lies on the ellipse x^2/b^2 +y^2/b^2 = 1. The normal at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = b^2/a^2.


help if you can
thanx

 

[spice girl]

Originally posted by gronkboyslim
P(a cos@, bsin @) lies on the ellipse x^2/b^2 +y^2/b^2 = 1. The normal at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = b^2/a^2.


help if you can
thanx

I'll just write down the method. It's just basically co-ordinate geometry.

1) Find the equation of the normal at P(whatever, whatever)

2) Find the points X, Y (remember all points on the x axis has y = 0, and all the points on the y axis has x = 0)

3) Find the distances of PX and PY (this is standard co-ord geometry)

4) Find PX / PY.