Skip to main content
Back to Conic Sections
JEE Main 2023
Conic Sections
Ellipse
Hard

Question

Let the ellipse E1:x2a2+y2b2=1E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a>ba > b and E2:x2A2+y2B2=1E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, A<BA < B have same eccentricity 13\frac{1}{\sqrt{3}}. Let the product of their lengths of latus rectums be 323\frac{32}{\sqrt{3}} and the distance between the foci of E1E_1 be 4. If E1E_1 and E2E_2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

Options

Solution

1. Introduction to Ellipse Properties

This problem involves two ellipses with different orientations but the same eccentricity. To solve it, we need to recall the key definitions and formulas for ellipses, including their standard equations, eccentricity, length of the latus rectum, and the distance between foci. We will then use these properties to determine the specific equations of both ellipses and subsequently find their intersection points to calculate the area of the quadrilateral formed by these points.

Let's define the parameters for each type of ellipse:

  • Horizontal Ellipse (E1E_1): The major axis lies along the x-axis.
    • Standard Equation: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where
Previous QuestionNext Question

Practice More Conic Sections Questions

View All Questions