Using the Lagrange's multiplier we need to solve the following problem.
At what point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ does the tangent line to it form with the coordinate axes a triangle of smallest area ?
I am stuck at the application of Lagrange's multiplier. Had it been some other way like coordinate geometry, I could have managed somehow. But here I have no idea how to proceed. Can someone help me out?
General tangent line to ellipse at point $(p,q)$ has equation
$$\frac{px}{a^2}+\frac{qy}{b^2}=1;\;\frac{p^2}{a^2}+\frac{q^2}{b^2}=1$$
(see here, for instance.)
Suppose $p,q$ positive. (The problem is symmetric)
Triangle area is $$A(p,q)=\frac{a^2b^2}{2pq};\;\frac{p^2}{a^2}+\frac{q^2}{b^2}=1$$
Define $$f(p,q,\lambda)=\frac{a^2b^2}{2pq}+\lambda\left(\frac{p^2}{a^2}+\frac{q^2}{b^2}-1\right)$$ $$ \begin{cases} \frac{2 \lambda p}{a^2}-\frac{a^2 b^2}{2 p^2 q}=0\\ \frac{2 \lambda q}{b^2}-\frac{a^2 b^2}{2 p q^2}=0\\ \frac{p^2}{a^2}+\frac{q^2}{b^2}=1\\ \end{cases} $$
solution is $p=\frac{a}{\sqrt{2}},q= \frac{b}{\sqrt{2}},\lambda=ab$.
Minimum area is $$A(p,q)=ab$$
Because of the symmetry the points are four $$\left(\pm\frac{a}{\sqrt{2}},\pm \frac{b}{\sqrt{2}}\right)$$
Hope this helps
