The centre of a circle C is at the centre of the ellipse E:x^2a^2+y^2b^2=1, ab.

Question

The centre of a circle $C$ is at the centre of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\ a>b$. Let $C$ pass through the foci $F_1$ and $F_2$ of $E$ such that the circle and the ellipse intersect at four points. Let $P$ be one of these four points. If the area of the triangle $PF_1F_2$ is $30$ and the length of the major axis of $E$ is $17$, then the distance between the foci of $E$ is:

Accepted Answer

$C$ has radius $c$ (it passes through the foci, centred at the ellipse's centre), so the intersection points $P$ have $x=\pm c$; their height satisfies $|y|=\dfrac{b^2}{\sqrt{a^2-b^2}}=\dfrac{b^2}{c}$ (from substituting $x=\pm c$ into the ellipse).

$$\text{Area}(PF_1F_2)=\frac12(2c)\left(\frac{b^2}{c}\right)=b^2=30.$$

With $2a=17$: $c^2=a^2-b^2=\dfrac{289}{4}-30=\dfrac{169}{4}\Rightarrow c=\dfrac{13}{2}$.

Distance between foci $=2c=13$.

Solutions · 1