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JEE Main 2024
Conic Sections
Hyperbola
Medium

Question

Consider the hyperbola x2a2y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 having one of its focus at P(3,0)\mathrm{P}(-3,0). If the latus ractum through its other focus subtends a right angle at P and a2b2=α2β,α,βNa^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}, then α+β\alpha+\beta is _________ .

Answer: 3

Solution

This solution will guide you through solving a problem involving hyperbolas, focusing on key properties like foci, eccentricity, latus rectum, and the condition for perpendicular lines.


1. Understanding the Standard Hyperbola and its Foci

  • Key Concept: For a hyperbola with the standard equation x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, where the transverse axis lies along the x-axis, the coordinates of its foci are (±ae,0)(\pm ae, 0). Here, aa is the semi-transverse axis length, bb is the semi-conjugate axis length, and ee is the eccentricity. The eccentricity ee is related to aa and bb

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