Ellipse
Ellipse
nta_pyq_2025_apr
Grade 11
Question:
Let the ellipse $E_1 : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b$ and $E_2 : \dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1$, $A < B$ have the same eccentricity $\dfrac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\dfrac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A$, $B$, $C$ and $D$, then the area of the quadrilateral $ABCD$ equals
$\dfrac{12\sqrt{6}}{5}$
$6\sqrt{6}$
$\dfrac{18\sqrt{6}}{5}$
$\dfrac{24\sqrt{6}}{5}$
Step-by-Step Solution
Key Concept: Determine $E_1$ from the foci distance and eccentricity; use the latus rectum product to find $E_2$; solve the two ellipse equations simultaneously to get the four intersection points forming a rectangle, then compute its area.
$2ae=4, e=\tfrac{1}{\sqrt{3}} \Rightarrow a=2\sqrt{3}$, $b^2=12(1-\tfrac{1}{3})=8$. $E_1: \tfrac{x^2}{12}+\tfrac{y^2}{8}=1$. Product of latus rectums: $\tfrac{2b^2}{a}\cdot\tfrac{2A^2}{B}=\tfrac{2\cdot8}{2\sqrt{3}}\cdot\tfrac{2A^2}{B}=\tfrac{32}{\sqrt{3}}$, giving $A^2/B=2$. Also $1-A^2/B^2=\tfrac{1}{3}$, so $A^2=\tfrac{2B^2}{3}$. Combining: $B=3$, $A^2=6$. $E_2:\tfrac{x^2}{6}+\tfrac{y^2}{9}=1$. Solving $E_1\cap E_2$: $y^2=\tfrac{36}{5}$, $x^2=\tfrac{6}{5}$. The four points form a rectangle with area $4\cdot\sqrt{\tfrac{6}{5}}\cdot\sqrt{\tfrac{36}{5}}=4\cdot\tfrac{6\sqrt{6}}{5}=\dfrac{24\sqrt{6}}{5}$.
Correct Answer: 4