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- Thread starter MarkFL
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- Feb 7, 2012

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Extremely clever, Chris, but then I would expect no less from you!

Here is my solution:

The equation of the ellipse is then:

(1) \(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

and the hypotenuse of the right triangle lies along the line:

(2) \(\displaystyle y=-\frac{h}{B}x+(h-b)\)

Substituting for $y$ from (2) into (1), we obtain the quadratic in $x$:

\(\displaystyle \left(\frac{a^2h^2}{B^2}+b^2 \right)x^2+\left(\frac{2a^2h(b-h)}{B} \right)x+a^2h(h-2b)=0\)

Since the ellipse is tangent to the hypotenuse, we require the discriminant to be zero, which implies:

\(\displaystyle h=\frac{2bB^2}{B^2-a^2}\)

And so the area $A$ of the right triangle may be written as as function of $B$ as follows:

\(\displaystyle A(B)=\frac{bB^3}{B^2-a^2}\)

Now, differentiating with respect to $B$ and equating the result to zero, we obtain:

\(\displaystyle A'(B)=\frac{bB^2\left(B^2-3a^2 \right)}{\left(B^2-a^2 \right)^2}=0\)

Since we must have \(\displaystyle a<B\) we find:

\(\displaystyle B^2=3a^2\)

and hence:

\(\displaystyle h=\frac{2b\left(3a^2 \right)}{3a^2-a^2}=3b\)

It is easy to see this minimizes the area of the triangle by the first derivative test.