Problem 832

## Solution.

The given curves are the parabola $y^{2}=2 p x$ and the ellipse $x^{2}+4 y^{2}=2 p^{2}$.

To determine the area of the desired triangle, we need to find the equations of the common tangents to the parabola and the ellipse. Their intersections with the coordinate axes determine the vertices $A$, $B$, and $C$ of the desired triangle.

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(A sketch of the curves with clearly marked tangents and vertices of the desired triangle or a shaded triangle is worth 2 points, and if the triangle is not marked, assign 1 point, as well as if the student only recognizes that they are dealing with a parabola and an ellipse.)

The equation of the ellipse can be written in the form $\frac{x^{2}}{2 p^{2}}+\frac{y^{2}}{\frac{p^{2}}{2}}=1$, from which we have $a^{2}=2 p^{2}, b^{2}=\frac{p^{2}}{2}$.

We use the condition that the line $y=k x+l$ is tangent to the parabola, so $p=2 k l$, and that it is tangent to the ellipse, so $k^{2} a^{2}+b^{2}=l^{2}$. Thus, we solve the following system of equations (with unknowns $k$ and $l$)

$$p=2 k l, \quad k^{2} \cdot 2 p^{2}+\frac{p^{2}}{2}=l^{2}$$

If we express $l=\frac{p}{2 k}$ from the first equation, then we get $2 p^{2} k^{2}+\frac{p^{2}}{2}=\frac{p^{2}}{4 k^{2}}$, which after dividing by $p^{2}$ and rearranging gives the biquadratic equation $8 k^{4}+2 k^{2}-1=0$. Since $k$ is a real number, it must be $k^{2} \geqslant 0$, i.e., $k^{2}=\frac{1}{4}$. Then $k= \pm \frac{1}{2}$, and from $p=2 k l$ we get $l= \pm p$.

(Since $p$ is a positive real number, $l$ and $k$ have the same sign.)

Thus, the equations of the tangents are

$$t_{1} \ldots y=\frac{1}{2} x+p \quad t_{2} \ldots y=-\frac{1}{2} x-p$$

The intersections of these tangents with the $y$-axis are the points $C(0, p)$ and $B(0,-p)$, and the intersection of the tangents (and the $x$-axis) is the point $A(-2 p, 0)$.

If we denote the origin by $O$, the area of the triangle $A B C$ is

$$P=\frac{|A O| \cdot|B C|}{2}=\frac{2 p \cdot 2 p}{2}=2 p^{2}$$

1 point