Lagrange Multipliers: find points farthest/closest to a point

Question

Find the points of the ellipse:

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

which are closest to and farthest from the point $(1,1)$.

I use the method of the Lagrange Multipliers by setting:

$$f(x,y)=(x-1)^2+(y-1)^2$$

(no need for the root since maximizing one maximizes the other)

And $$g(x,y)=\frac{x^2}{9}+\frac{y^2}{4}-1=0$$ the constraint.

We get:

$$x-1=λx/9$$ and $$2y-2=λy/2$$

I don't know where to go from there. I tried finding $x$ and $y$ alone and then plugging their values in the constraint $g(x,y)=0$ but I get a really complicated equation.

Answer 1

Hint: $$\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y)=(x-1)^2+(y-1)^2+\lambda \left(\frac{x^2}{9}+\frac{y^2}{4}-1\right) $$

Then:

$\frac{\partial \mathcal{L}}{\partial x}=2(x-1)+\frac{2x\lambda}{9}=0 \implies x=\frac{9}{9+\lambda}$

$\frac{\partial \mathcal{L}}{\partial y}=2(y-1)+\frac{\lambda y}{2}=0 \implies y=\frac{4}{4+\lambda}$

$\frac{\partial \mathcal{L}}{\partial \lambda}= \frac{x^2}{9}+\frac{y^2}{4}-1=0$

so $\implies \frac{\left(\frac{9}{9+\lambda}\right)^2}{9}+\frac{\left(\frac{4}{4+\lambda}\right)^2}{4}=1$

$\implies \frac{9}{(9+\lambda)^2}+\frac{4}{(4+\lambda)^2}=1$

Then solve for $\lambda$ (not easy) then plug in value to obtain $x=\frac{9}{9+\lambda}$ and $y=\frac{4}{4+\lambda}$ that are required

Answer 2

HINT:

$$ f(x,y)=(x-1)^2+(y-1)^2 $$

$$ g(x,y)= \left(\frac{x^2}{9}+\frac{y^2}{4 } \right) $$

$$ -\lambda =\dfrac{ \dfrac{\partial f }{\partial x } }{\dfrac{\partial g }{\partial x } } =\dfrac{ \dfrac{\partial f }{\partial y } }{\dfrac{\partial g }{\partial y } }. $$