Find a length of base and leg of minimum area isosceles triangle circumscribed about a ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ if its base is parallel to x-axis
Similar problem, where I am given circle instead of ellipse is easy, because I can use similar triangles to write leg in terms of base and diameter of circle (or vice versa), but in this case I have no idea because I don't know what's the distance from center of ellipse to leg of triangle.
I thought about writing an eqation of ellipse in parametric form but that didn't seem to help me much.
How should I solve this problem?
Best Answer
In the figure below, let $F=(0,h)$ and $E=(x_0,y_0)$, with $FE$ tangent to the ellipse.
Then $$\frac{x_0^2}4+\frac{y_0^2}9=1$$ The slope of the tangent is given by the derivative: $$y'=\left(3\sqrt{1-\frac{x^2}4}\right)'=-\frac34\frac x{\sqrt{1-\frac{x^2}4}}=-\frac94\frac xy$$ You could have gotten this result with using implicit derivative.
Then, at $E$ the slope of the tangent is equal to the slope of the $FE$ line.
$$\frac{y_0-h}{x_0}=-\frac94\frac{x_0}{y_0}$$ Rearranging the terms: $$\frac{y_0^2}9-\frac{y_0h}9+\frac{x_0^2}4=0$$ or $$y_0=\frac9h$$ From here, calculate $x_0$ in terms of $h$, then the $x$ coordinate of $G$ (knowing that the $y$ coordinate is $-3$). Then the area of the triangle is $x_G(h)\times (h+3)$. Take the derivative, set it to zero, and you get $h$.