Let C be the circle of minimum area enclosing the ellipse E:x^2a^2+y^2b^2=1 with eccentric…

Question

Let $C$ be the circle of minimum area enclosing the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with eccentricity $\dfrac12$ and foci $(\pm2,0)$. Let $PQR$ be a variable triangle whose vertex $P$ is on the circle $C$, and the side $QR$ of length $2a$ is parallel to the major axis of $E$ and contains the point where $E$ meets the negative $y$-axis. Then the maximum area of the triangle $PQR$ is:

Accepted Answer

$ae=2$ with $e=\dfrac12\Rightarrow a=4$; $b^2=a^2(1-e^2)=12\Rightarrow b=2\sqrt3$. The minimum enclosing circle $C$ has radius $a=4$, centred at the origin.

$QR$ is the horizontal segment of length $2a$ through $(0,-b)$. The farthest point of $C$ from this line is the top of the circle, at height $a+b$ above it (since $P$ can reach $y=a$).

$$\text{Max Area}=\frac12\times2a\times(a+b)=a(a+b)=4(4+2\sqrt3)=8(2+\sqrt3).$$

Solutions · 1