# points on ellipse closest to a point

• Jun 25th 2012, 04:32 AM
khh
points on ellipse closest to a point
Hi

Find the points of the ellipse 4x2 + 9y2 =36 that are closest to the point (1,1), as well as the point or points farthest from it?

I have done so far:
g(x,y)=4x2 + 9y2 - 36 = 0 ==> gx(x,y)=8x & gy(x,y)=18y

The distance from the point is d=sqrt[(x-1)2 + (y-1)2 ]
==> f(x,y)= (x-1)2 + (y-1)2

==> fx(x,y)=
λ.gx(x,y) and fy(x,y)=λ.gy(x,y)

so 2(x-1) = λ . 8x and 2(y-1) = λ . 18y and from above 4x2 + 9y2 - 36 = 0

How do I solve this?

• Jun 27th 2012, 11:04 PM
a tutor
Re: points on ellipse closest to a point
From $\displaystyle 2(x-1)=8 \lambda x$ you can get $\displaystyle \lambda = \frac{x-1}{4x}$.

Similarly you can get lambda in terms of y. Equate these to get an equation that will allow you to find x in terms of y or y in terms of x.

This is where I stopped. $\displaystyle x=\frac{9y}{5y+4}$ unless I messed up.

Sub this in your ellipse equation and solve.
• Jun 28th 2012, 02:03 AM
a tutor
Re: points on ellipse closest to a point
Back again. :)

That leads to a quartic equation which I needed my calculator to solve. Once you have your y values find the x values and don't forget to check the distance for each point you find.
• Jun 29th 2012, 06:04 AM
a tutor
Re: points on ellipse closest to a point
OK. So I got your PM.

If you got as far as the quartic equation why didn't you say so?

I don't have it written anywhere now but it's easy enough to solve, using the Newton Raphson method.

Post it if you like and I'll check it.

Do you know the Newton Raphson method?