Skip to main content
Back to Conic Sections
JEE Main 2020
Conic Sections
Hyperbola
Medium

Question

For the hyperbola H:x2−y2=1\mathrm{H}: x^{2}-y^{2}=1 and the ellipse E:x2a2+y2 b2=1\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1, a >b>0>\mathrm{b}>0, let the (1) eccentricity of E\mathrm{E} be reciprocal of the eccentricity of H\mathrm{H}, and (2) the line y=52x+Ky=\sqrt{\frac{5}{2}} x+\mathrm{K} be a common tangent of E\mathrm{E} and H\mathrm{H}. Then 4(a2+b2)4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right) is equal to _____________.

Answer: 2

Solution

This problem requires a strong understanding of the properties of hyperbolas and ellipses, specifically their eccentricities and conditions for tangency. We will systematically use the given information to establish relationships between the parameters of the ellipse and then solve for the required value.

1. Understand the Equations and Identify Parameters

First, let's clearly state the standard forms and identify the parameters for the given conics.

  • Hyperbola H: The given equation is x2−y2=1x^2 - y^2 = 1. Comparing this with the standard form of a hyperbola centered at the origin, x2A2−y2B2=1\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1, we can identify its parameters:
Previous QuestionNext Question

Practice More Conic Sections Questions

View All Questions