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JEE Main 2023
Conic Sections
Hyperbola
Hard

Question

Let e 1 and e 2 be the eccentricities of the ellipse x2b2+y225=1\frac{x^2}{b^2} + \frac{y^2}{25} = 1 and the hyperbola x216y2b2=1\frac{x^2}{16} - \frac{y^2}{b^2} = 1, respectively. If b < 5 and e 1 e 2 = 1 , then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

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Solution

This problem is an excellent test of your understanding of conic sections, specifically ellipses and hyperbolas, their eccentricities, and the locations of their foci. We will systematically break down the problem, calculate the necessary parameters, and finally determine the eccentricity of the new ellipse.


1. Understanding the Properties of the First Ellipse

The first ellipse is given by the equation x2b2+y225=1\frac{x^2}{b^2} + \frac{y^2}{25} = 1. Let its semi-major axis be A1A_1 and semi-minor axis be B1B_1. From the equation, we have A12=25A_1^2 = 25 and B12=b2B_1^2 = b^2 (or vice versa).

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