Hyperbola
Hyperbola
nta_pyq_2025_apr
Grade 11
Question:
Let $E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b$ and $H : \dfrac{x^2}{A^2} - \dfrac{y^2}{B^2} = 1$. Let the distance between the foci of $E$ and the foci of $H$ both be $2\sqrt{3}$. If $a - A = 2$, and the ratio of the eccentricities of $E$ and $H$ is $\tfrac{1}{3}$, then the sum of the lengths of their latus rectums is equal to
Step-by-Step Solution
Key Concept: Both conics share the same focal distance $2\sqrt{3}$, so $ae = Ae' = \sqrt{3}$; use $e/e'=1/3$ and $a-A=2$ to find $a$, $A$, then $b^2$ and $B^2$; add the two latus rectum lengths.
$ae=\sqrt{3}$ and $Ae'=\sqrt{3}$, so $ae=Ae' \Rightarrow \tfrac{e}{e'}=\tfrac{A}{a}=\tfrac{1}{3} \Rightarrow a=3A$. With $a-A=2$: $A=1$, $a=3$. $e=\tfrac{\sqrt{3}}{3}=\tfrac{1}{\sqrt{3}}$, $e'=\sqrt{3}$. $b^2=a^2(1-e^2)=9(\tfrac{2}{3})=6$. $B^2=A^2(e'^2-1)=1(2)=2$. Sum of LR $=\dfrac{2b^2}{a}+\dfrac{2B^2}{A}=\dfrac{12}{3}+\dfrac{4}{1}=4+4=8$.
Correct Answer: 3