Minimize area of ellipse that passes through given point The ellipse formula is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
$a$ and $b$ are assumed to be positive.
The point $(6, -7)$ is given as a point on the ellipse, with the area of the ellipse equal to $\pi ab$. 
How would one find the values of $a$ and $b$ that minimize the area of the ellipse?

Express $a$ in terms of $b$ given the equation of an ellipse and the given point:
You should get:
$$a=\frac{6b}{\sqrt{b^2-49}}$$
You want to maximize the area ($\pi a b$):
$$A=6\pi \cdot \frac{b^2}{\sqrt{b^2-49}}$$
Now differentiate $A$ with respect to $b$ and set $A'=0$ to find the critical points:
$$0=6\pi \left( \frac{2b\sqrt{b^2-49}-\frac{b^3}{\sqrt{b^2-49}}}{b^2-49}\right)$$
$$0= b\left(b^2-98\right)$$
Therefore, $b=0$ and $b= \pm 7\sqrt{2}$ are critical points, but $b \neq 0 \; \text{and} \; b \neq -7\sqrt{2}$.  Try finishing the rest.
