Ellipse - PLEASE HELP!!! (1 Viewer)

[Sunyata]

Hi

How exactly would I do this?

The normal to the ellipse x^2/a^2 + y^2/b^2 = 1 at P(x1, y1) meets the x-axis in N and the Y-axis in G. Prove PN/NG = (1-e^2)/e^2<!-- google_ad_section_end -->

 

[Mature Lamb]

The normal to the ellipse x^2/a^2 + y^2/b^2 = 1 at P(x1, y1) meets the x-axis in N and the Y-axis in G. Prove PN/NG = (1-e^2)/e^2

1. get equation of normal to the ellipse at P

- differentiate the equation of the ellipse with respect to x, then negative reciprocal of dy/dx for gradient of normal, then point gradient formula with P(x1, y1) to end up with:

a^2x / x1 - b^2y / y1 = a^2 - b^2




2. solve for x = 0 and y = 0 for points N and G

when y = 0, x = x1(a^2-b^2)/a^2

.'. N (x1(a^2-b^2)/a^2, 0)

when x = 0, y = -y1(a^2-b^2)/b^2

.'. G (0, -y1(a^2-b^2)/b^2)




3. P, N and G are on the same line, so..

using division of a line with a given ratio formula, prove that N divides line segment PG in ratio (1-e^2) : e^2 by showing the coordinates of N

NOTE:

from b^2 = a^2(1-e^2),

e^2 = (a^2 - b^2) / a^2



xn = (e^2.x1 + (1-e^2).0) / (1-e^2) - e^2

= e^2.x1

= x1(a^2-b^2) / a^2

which is the x co-ordinate of N

Do the same for the y-coordinate of N and prove that it's equal to 0.


P.S. sorry for the messy solutions.. I don't know how to work LaTeX equation editor

 

[cutemouse]

Use similiar triangles to make it less messy.

 

[Sunyata]

Ummm the y-coordiante doesnt seem to = 0

 

[js992]

There's a faster and less messy solution.

You want to prove PN/NG = (1-e^2)/e^2

Find your coordinate G
which is (0, [-y1(a^2-b^2)]/b^2)
Let B be the foot of the perpendicular for P to the y axis
So , B is (0,y1)

Draw it out and you can see that PN/NG = BO / BG by ratio of intercepts on a parallel line.

It becomes much simpler now as you dont need to use distance formula.

BO = y1
OG = [y1 (a^2-b^2)]/b^2

BO / OG = b^2/(a^2-b^2)
= b^2/a^2e^2 <=== (e^2 = 1-b^2/a^2)
= (b^2/a^2) x (1/e^2)
= (1-e^2) / e^2

Sorry its hard to read, latex isnt working =\

 

[Mature Lamb]

Use similiar triangles to make it less messy.

What are the triangles involved?

 

[cutemouse]

What are the triangles involved?

I'll post my solution tomorrow. It's way too late for that now...