arthham

JEE Advanced 2022 · Conic Sections

Consider the hyperbola x2100−y264=1\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1 with foci at SS and S1S_{1}, where SS lies on the positive xx-axis. Let PP be a point on the hyperbola, in the first quadrant. Let ∠SPS1=α\angle SPS_{1} = \alpha, with α<π2\alpha < \frac{\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 S1PS_{1}P at P1P_{1}. Let δ\delta be the distance of PP from the straight line SP1SP_{1}, and β=S1P\beta = S_{1}P. Then the greatest integer less than or equal to βδ9sin⁡α2\frac{\beta\delta}{9}\sin\frac{\alpha}{2} is ______.

Answer: 77

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Consider the hyperbola x²/100 - y²/64 = 1 with foci at S and S₁, where S lies on the positive x-axis. Let P be a point on the hyperbola, in the first quadrant. Let ∠ SPS₁ = α, with α < π/2. The straight line passing through the point S and having the same slope as that of the tangent at P to the hyperbola, intersects the straight line S₁P at P₁. Let δ be the distance of P from the straight line SP₁, and β = S₁P. Then the greatest integer less tha x2100−y264=1:a=10, b=8, c=241\dfrac{x^2}{100}-\dfrac{y^2}{64}=1:\quad a=10,\ b=8,\ c=2\sqrt{41} S(241,0),S1(−241,0),P in first quadrantS(2\sqrt{41},0),\quad S_1(-2\sqrt{41},0),\quad P \text{ in first quadrant} ∠SPS1=α<π2\angle SPS_1 = \alpha < \dfrac{\pi}{2} Line through S with slope=tangent slope at P\text{Line through } S \text{ with slope} = \text{tangent slope at } P meets S1P at P1;δ=dist(P, SP1), β=S1P\text{meets } S_1P \text{ at } P_1;\quad \delta=\text{dist}(P,\ SP_1),\ \beta=S_1P Find ⌊βδ9sin⁡α2⌋\text{Find } \left\lfloor \dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2} \right\rfloor Plan: 1. draw hyperbola, foci\text{Plan: } 1.\ \text{draw hyperbola, foci} 2. tangent at P; line through S⇒P12.\ \text{tangent at } P;\ \text{line through } S \Rightarrow P_1 3. use PS1−PS=2a and law of cosines3.\ \text{use } P S_1 - PS = 2a \text{ and law of cosines} x2100−y264=1,a=10, b=8\dfrac{x^2}{100}-\dfrac{y^2}{64}=1,\quad a=10,\ b=8 c=a2+b2=164=241c=\sqrt{a^2+b^2}=\sqrt{164}=2\sqrt{41} S(241,0),S1(−241,0)S(2\sqrt{41},0),\quad S_1(-2\sqrt{41},0) P on the hyperbola, first quadrantP \text{ on the hyperbola, first quadrant} ∠SPS1=α<π2\angle SPS_1=\alpha<\dfrac{\pi}{2} Tangent at P(x0,y0): xx0100−yy064=1\text{Tangent at } P(x_0,y_0):\ \dfrac{xx_0}{100}-\dfrac{yy_0}{64}=1 slope m=16x025y0\text{slope } m=\dfrac{16x_0}{25y_0} through S with slope m:SP1∥tangent at P\text{through } S \text{ with slope } m:\quad SP_1 \parallel \text{tangent at } P SP1∩S1P=P1SP_1 \cap S_1P = P_1 tangent at P bisects ∠SPS1\text{tangent at } P \text{ bisects } \angle SPS_1 δ=dist(P,SP1)=PS⋅sin⁡α2\delta = \text{dist}(P,SP_1)=PS\cdot\sin\dfrac{\alpha}{2} β=S1P\beta=S_1P d2−d1=2a=20,d1=SP, d2=S1Pd_2-d_1=2a=20,\quad d_1=SP,\ d_2=S_1P SS12=d12+d22−2d1d2cos⁡α,SS1=2cSS_1^2=d_1^2+d_2^2-2d_1d_2\cos\alpha,\quad SS_1=2c 4c2=(d2−d1)2+2d1d2(1−cos⁡α)4c^2=(d_2-d_1)^2+2d_1d_2(1-\cos\alpha) 4c2=4a2+4d1d2sin⁡2α24c^2=4a^2+4d_1d_2\sin^2\dfrac{\alpha}{2} d1d2sin⁡2α2=c2−a2=b2=64d_1d_2\sin^2\dfrac{\alpha}{2}=c^2-a^2=b^2=64 βδsin⁡α2=d1d2sin⁡2α2=64\beta\delta\sin\dfrac{\alpha}{2}=d_1d_2\sin^2\dfrac{\alpha}{2}=64 βδ9sin⁡α2=649≈7.11\dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2}=\dfrac{64}{9}\approx 7.11 ⌊βδ9sin⁡α2⌋=7\left\lfloor \dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2} \right\rfloor = 7

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