Ellipse
Ellipse
nta_pyq_2025_apr
Grade 11
Question:
Let the product of the focal distances of the point $\left(\sqrt{3}, \tfrac{1}{2}\right)$ on the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $(a > b)$, be $\tfrac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is
$\dfrac{1 - \sqrt{3}}{\sqrt{2}}$
$\dfrac{3 - 2\sqrt{2}}{2\sqrt{3}}$
$\dfrac{3 - 2\sqrt{2}}{3\sqrt{2}}$
$\dfrac{1 - 2\sqrt{2}}{\sqrt{3}}$
Step-by-Step Solution
Key Concept: Product of focal distances $= a^2 - e^2x_1^2$; the point also lies on the ellipse giving a second equation in $a^2$ and $e^2$; eliminate $a^2$ to get a quadratic in $e^2$ with two valid roots.
Product of focal distances $=(a+ex_1)(a-ex_1)=a^2-3e^2=\tfrac{7}{4}$, so $a^2=\tfrac{7}{4}+3e^2$. The point lies on the ellipse: $\tfrac{3}{a^2}+\tfrac{1}{4b^2}=1$, with $b^2=a^2(1-e^2)$. Substituting and simplifying gives $12e^4-17e^2+6=0$, so $e^2=\tfrac{3}{4}$ or $\tfrac{2}{3}$, giving $e=\tfrac{\sqrt{3}}{2}$ or $\sqrt{\tfrac{2}{3}}$. Absolute difference $=\tfrac{\sqrt{3}}{2}-\sqrt{\tfrac{2}{3}}=\dfrac{3-2\sqrt{2}}{2\sqrt{3}}$.
Correct Answer: 2