A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse, two curves intersect in $4$ points.

Question

A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse,such that the two curves intersect in $4$ points.Let $P$ be any one of their point of intersection.If the major axis of the ellipse is $17$ and the area of the triangle $PF_1F_2$ is $30$,then find the distance between the foci.

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Let the center of the ellipse and the circle be $(0,0)$

We are given $2a=$length of major axis$=17$.

Let the coordinates of foci be $F_1(c,0)$ and $F_2(-c,0)$

We need to find $2c$.
Area of $PF_1F_2=\frac{1}{2}\times 2c\times$perpendicular distance between $P$ and the axis of the major axis of the ellipse.

I do not know how to solve it further.

Answer 1

We may suppose that $$\text{the ellipse$\ :\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a\gt b\gt 0$}$$ $$\text{the circle$\ :\ x^2+y^2=a^2-b^2$}$$

As you wrote, we have $$2a=17\quad\Rightarrow \quad a=\frac{17}{2}$$

Since $$\frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\quad\Rightarrow\quad |y|=\frac{b^2}{\sqrt{a^2-b^2}}$$

we have $$30=\frac 12\times 2\sqrt{a^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}\quad\Rightarrow\quad b=\sqrt{30}.$$

Thus, the answer is $$2\sqrt{a^2-b^2}=2\sqrt{\left(\frac{17}{2}\right)^2-30}=\color{red}{13}.$$

Answer 2

HINT:

WLOG let the equations of the two curves be

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2(1-e^2)}=1$$

and $$x^2+y^2=b^2$$

Then $b=ae$ and $F_1,F_2$ are $(\pm ae,0)$

If $P(ae\cos A,ae\sin A),$

$$e^2\cos^2A+\dfrac{e^2\sin^2A}{(1-e^2)}=1$$

$$\iff e^4\cos^2A-2e^2+1=0$$

Find $e^2$ in terms of $A$

Can you take it from here?