Identifying the simplest method (1 Viewer)

[Yianni]

Hello all
Recently I came across the following question:
edit: This is a Conics Question for those who have not done it

P(acos@), bsin@) lies on the ellipse (x^2/a^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.

(From exercise 3.3 q4 From Cambridge)

The ways in which I approached this method included:
Simply finding distance : overly tedious and ugly
Interval division : Dividing PY by x and finding the ratio
Somehow using Triangles formed from tangent & normal intersection with the x/y axis.

I cant identify any other ways of approaching this question, and the ones above are not too efficient and very lengthy.

How would you do it?
(Im teaching myself conics; im not asking for an answer (at this point!), but if you have any thoughts i would appreciate them.
Ill try again tomorrow morning, and ill post what i get if anyone is interested.)

Thanks for your time.

 

[Affinity]

Somehow using Triangles formed from tangent & normal intersection with the x/y axis.

This takes around 1-2 mins.. which isn't too bad

the slope of the tangent is M

so the tangent is (x-acos(t))*M = y- bsin(t)
the y intercept is at x = 0 which turns out to be (0, ?)

PX / PY = bsin(t) / (bsin(t) - ?)
=... after some crunching b^2/a^2

 

[thunderdax]

with some questions like that you just have to crunch them out...

 

[Estel]

heh I usually go for triangles with qs like that (not that I have a choice) since my algebra is abysmal.

 

[ngai]

How would you do it?

i would use triangles/geometry/whatever you call it...like this:

draw a vertical line through P, and have that meet x-axis at A
need to find PX/PY
we can instead find PY/PX then flip it over
but PY/PX = (PX+XY)/PX = 1 + (XY/PX)
and XY/PX = OY/PA by similar triangles
OY is just |y_Y|, which is easily found by quoting the normal eqn and sub x=0
PA is just |y_P|, which is given