A circle has the same centre as an ellipse & passes through the foci FI & F2 of the ellipse, such that the two curves interseet in 4 points. Let 'P' be any one of their point of intersection. If the major axis of the ellipse is 17; the area of the triangle PFlF2 is 30, then the distance between the foci is:

I just know that 2a=17

after that I am not getting any idea of using the are information


marked as duplicate by David K, amd, Joel Reyes Noche, user91500, Namaste May 1 '17 at 12:04

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We do not need to solve five constraints. Three will do.

The circle passes through $F_1$ and $F_2$ with the center directly between them. So $\angle F_1PF_2$ with $P$ on the circle measures $90°$, therefore

$|PF_1|^2+|PF_2|^2=|F_1F_2|^2............(1) $

$P$ also lies in the ellipse whose major axis is given and must be the sum of the lengths from the foci to $P$:

$|PF_1|+|PF_2|=17............(2) $

The given area of $\triangle F_1PF_2$ which is a right triangle must be half the product of the legs:

$|PF_1|\times|PF_2|=60............(3) $

Square both sides of (2), and on the squared left side plug in the familiar binomial formula for the square of a sum. Use (3) to eliminate the cross product term and now you can get $|PF_1|^2+|PF_2|^2$ for use in Equation (1).

Not unlucky at all (subtle hint at the answer), you should find that $|F_1F_2|^2$ is a perfect square and thus $|F_1F_2|$ is a whole number. In fact, although you don't need it, $|PF_1|$ and $|PF_2|$ wind up being whole numbers too.

  • 1
    $\begingroup$ [+1] Yes, this way to consider the problem is definitely more in the spirit of the question. My solution, although rather efficient using a CAS, looks more "brute force" compared to yours... $\endgroup$ – Jean Marie Apr 30 '17 at 7:11

I use hereafter the classical notations $a,b$ for the semi axes, and $f:=OF_1=OF_2$.

I denote by $(x_0,y_0)$ the coordinates of $P$.

We have $five$ unknowns $a,b,f,x_0,y_0$ (I consider $a$ as an unknown although it is known, but it's more practical for setting formulas) and $five$ constraints:

$$\begin{cases}(1) & 2a=17 & \text{(given)}\\ (2) & a^2=b^2+f^2 & \text{(classical relationship for ellipses)}\\ (3) & x_0^2+y_0^2=f^2 & \text{($P$ belongs to the circle with radius $f$)}\\ (4) & \tfrac{x_0^2}{a^2}+\tfrac{y_0^2}{b^2}=1 & \text{($P$ belongs to the ellipse)}\\ (5) & fy_0=30 & \text{(area of triangle: half base $\times$ height $= \tfrac12 2f \times y_0$)} \end{cases}$$

I let you finish...

You should find $b=\sqrt{30}, \ f=\tfrac{13}{2}$, with $x_0=\tfrac{119}{26}, \ y_0=\tfrac{120}{26}.$

  • $\begingroup$ I have added the values you should obtain. $\endgroup$ – Jean Marie Apr 29 '17 at 9:15

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