Find the points closest and farthest on 4x^2 + 9y^2 = 36 to (1,1).
The systems is
2(x-1) = l (8x)
2(y-1) = l (18y)
4x^2 + 9y^2 -36=0
If I solve for x and y and substitute it into the last equation, I get a quartic equation in terms of lamda. If I solve for lambda in the 2 first equations, I get a cross term xy. How do I solve this?
You have $\lambda = \frac{2y-2}{18y} = \frac{2x-2}{8x}$ so $16xy - 16x = 36xy -36y$ implies $16x + 20xy = 36y$, hence $x = \frac{36y}{16 + 20y}$.
Now plugging into our equation 3, $4(\frac{36y}{16+20y})^2 + 9y^2 = 36 = \frac{5184y^2}{256 + 640y + 400y^2} + 9y^2$, so $ 5184y^2 + 2304y^2 + 5760y^3 + 3600y^4 = 9216 + 23040y + 14400y^2. $
Can you take it from here?