Skip to main content
Back to Conic Sections
JEE Main 2023
Conic Sections
Hyperbola
Easy

Question

Let the hyperbola H:x2a2y2=1H:{{{x^2}} \over {{a^2}}} - {y^2} = 1 and the ellipse E:3x2+4y2=12E:3{x^2} + 4{y^2} = 12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH{e_H} and eE{e_E} are the eccentricities of H and E respectively, then the value of 12(eH2+eE2)12\left( {e_H^2 + e_E^2} \right) is equal to ___________.

Answer: 2

Solution

This solution will guide you through the process of solving the given problem by systematically applying the standard formulas for hyperbolas and ellipses. We will focus on clarity, detailed explanations, and proper mathematical notation.


1. Understanding the Problem and Key Concepts

The problem asks us to calculate the value of 12(eH2+eE2)12(e_H^2 + e_E^2), where eHe_H and eEe_E are the eccentricities of a given hyperbola HH and an ellipse EE, respectively. A crucial piece of information is that the length of the latus rectum of HH is equal to the length of the latus rectum of EE.

To solve this, we need to recall the standard forms and associated formulas for hyperbolas and ellipses, specifically

Practice More Conic Sections Questions

View All Questions