Home › MCQs › JEE MAIN Mathematics › Question #981
Back to Questions
JEE MAIN Mathematics QUESTION #981
Question 1
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 \(H\) be \(2\sqrt{3}\). If \(a-A=2\) and the ratio of eccentricities of \(E\) and \(H\) is \(\dfrac{1}{3}\), then the sum of the lengths of their latus rectums is equal to:
  • 10
  • 9
  • 8✔️
  • 7
Correct Answer Explanation
Let \(e_E=e\) and \(e_H=3e\). Focal distances: \(2ae=2\sqrt{3}\Rightarrow ae=\sqrt{3}\) and \(2Ae_H=2\sqrt{3}\Rightarrow 3Ae=\sqrt{3}\Rightarrow Ae=\frac{1}{\sqrt{3}}\). Then \(\frac{a}{A}=3\) and \(a-A=2\Rightarrow A=1, a=3\). So \(e=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\), \(b^2=a^2(1-e^2)=9\cdot\frac{2}{3}=6\). For \(H\): \(e_H=\sqrt{3}\), \(B^2=A^2(e_H^2-1)=2\). Sum of latus rectums \(=\frac{2b^2}{a}+\frac{2B^2}{A}=\frac{12}{3}+\frac{4}{1}=4+4=8\).