points on ellipse closest to a point

Hi

Can someone please help me with the following problem?

Find the points of the ellipse 4x^{2 }+ 9y^{2 }=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)=4x^{2 }+ 9y^{2 }- 36 = 0 ==> g_{x}(x,y)=8x & g_{y}(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}

==> f_{x}(x,y)=λ.g_{x}(x,y) and f_{y}(x,y)=λ.g_{y}(x,y)

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

How do I solve this?

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.

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.

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?