What is the shortest/longest distance from $9x^2 + 4y^2 = 36$ to $(5,5)$?

Question

What is the shortest/longest distance from $9x^2 + 4y^2 = 36$ to $(5,5)$?

Using Langrange Multipliers, I've set up the standard equation with $$g(x,y) = (x/2)^2 + (y/3)^2 = 1$$ $$f(x,y) = (x-5)^2 + (y-5)^2$$ and $$ \nabla f = -\lambda \nabla g.$$ This gives us $$2(x-5) = - \lambda x / 2$$ $$(y-5) = - \lambda y / 9.$$ Solving for $y$ and $x$, I have $$ y = 9x / (x + 4) $$ $$ x = 4y / (9-y) .$$ But if I plug this value for $y$ into the original ellipse, I get $$ x^4 + 8x^3 + 48x^2 - 32x - 64 = 0.$$ Somehow this doesn't seem quite right as it's now cumbersome to solve for $x$. Where am I going wrong and is there a better approach?

Answer 1

We have $$\frac{x^2}4+\frac{y^2}9=1$$

so, any point can be written as $(2\cos t,3\sin t)$

If the distance if $d,$

$$d^2=(2\cos t-5)^2+(3\sin t-5)^2$$

Now use Second derivative test