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JEE Main 2024
Conic Sections
Hyperbola
Hard

Question

Let m1m_{1} and m2m_{2} be the slopes of the tangents drawn from the point P(4,1)\mathrm{P}(4,1) to the hyperbola H:y225x216=1H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1. If Q\mathrm{Q} is the point from which the tangents drawn to H\mathrm{H} have slopes m1\left|m_{1}\right| and m2\left|m_{2}\right| and they make positive intercepts α\alpha and β\beta on the xx-axis, then (PQ)2αβ\frac{(P Q)^{2}}{\alpha \beta} is equal to __________.

Answer: 2

Solution

This problem combines the concept of tangents to a hyperbola, specifically using the slope form, with careful analysis of x-intercepts and coordinate geometry. We will break down the solution into clear, manageable steps.

Key Concept: Tangent Equation of a Hyperbola in Slope Form For a hyperbola of the form y2a2x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 (transverse axis along the y-axis), the equation of a tangent with slope mm is given by: y=mx±a2m2b2y = mx \pm \sqrt{a^2m^2 - b^2} This formula is critical for solving the problem. Note that the sign before the square root depends on the specific tangent and

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