Let E:x^216+y^29=1.

Question

Let $E:\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$. For two distinct values of $p$, the lines $y=x+p$ touch the ellipse $E$ at the points $A$ and $B$. Let the line $y=x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to:

Accepted Answer

Tangency: $p^2=a^2m^2+b^2=16(1)+9=25\Rightarrow p=\pm5$.

Points of contact ($x_1=-a^2m/c,\ y_1=b^2/c$ for tangent $y=mx+c$): $A=\left(-\dfrac{16}{5},\dfrac95\right)$ (for $p=5$), $B=\left(\dfrac{16}{5},-\dfrac95\right)$ (for $p=-5$).

$y=x$ meets $E$: $x^2\left(\dfrac1{16}+\dfrac19\right)=1\Rightarrow x=\pm\dfrac{12}{5}$, giving $C=\left(\dfrac{12}{5},\dfrac{12}{5}\right),\ D=\left(-\dfrac{12}{5},-\dfrac{12}{5}\right)$.

Since $B=-A$ and $D=-C$, $AB$ and $CD$ are both diagonals through the origin, so $ACBD$ is a parallelogram. Using the shoelace formula on $A,C,B,D$ (in that cyclic order):

$$\text{Area}=24.$$

Solutions · 1