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JEE Advanced 2022Paper 2
Exam
JEE Advanced
Year
2022
Paper
Paper 2
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2022 · Paper 2 · Mathematics · Hyperbola

Consider the hyperbola x ^ 2 100 - y ^ 2 64 = 1 with foci at S and...

Consider the hyperbola x2100−y264=1\dfrac{{x}^{2}}{100} - \dfrac{{y}^{2}}{64} = 1 with foci at SS and S1{S}_{1}, where SS lies on the positive xx-axis. Let PP be a point on the hyperbola, in the first quadrant. Let ∠SPS1=α∠ SP{S}_{1} = \alpha, with α<π2\alpha < \dfrac{\pi}{2}. The straight line passing through the point SS and having the same slope as that of the tangent at PP to the hyperbola, intersects the straight line S1P{S}_{1}P at P1{P}_{1}. Let δ\delta be the distance of PP from the straight line SP1S{P}_{1}, and β=S1P\beta = {S}_{1}P. Then the greatest integer less than or equal to βδ9sinα2\dfrac{\beta\delta}{9}sin\dfrac{\alpha}{2} is _____ .Consider the hyperbola x ^ 2 100 - y ^ 2 64 = 1 with foci at S and S _ 1 , where S lies on the positive x -axis. Let P

Correct answer

0

Step-by-step solution

Plotting the diagram of the given value we have,

Consider the hyperbola x ^ 2 100 - y ^ 2 64 = 1 with foci at S and S _ 1 , where S lies on the positive x -axis. Let P

We know that product of distances of any tangent from two foci =b2= {b}^{2}

So, δ×βsinα2=b2\delta \times \beta sin\dfrac{\alpha}{2} = {b}^{2}

⇒βδsinα29=b29=649⇒ \dfrac{\beta\delta sin\dfrac{\alpha}{2}}{9} = \dfrac{{b}^{2}}{9} = \dfrac{64}{9}

Now finding the greatest integer less than or equal to βδsinα29\dfrac{\beta\delta sin\dfrac{\alpha}{2}}{9} which will be 77

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