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MathematicsInteger+4 / -1Question 59 of 60

JEE Advanced 2013 · Paper 1, Mathematics question 59

A vertical line passing through the point (h,0)(h,0) intersects the ellipse x24+y23=1\dfrac{x^2}4+\dfrac{y^2}3=1 at the points PP and QQ. Let the tangents to the ellipse at PP and QQ meet at the point RR. If Δ(h)=\Delta(h)= area of the triangle PQRPQR, Δ1=max⁡1/2≤h≤1Δ(h)\Delta_1=\displaystyle\max_{1/2\le h\le1}\Delta(h) and Δ2=min⁡1/2≤h≤1Δ(h)\Delta_2=\displaystyle\min_{1/2\le h\le1}\Delta(h), then 85Δ1−8Δ2=\dfrac8{\sqrt5}\Delta_1-8\Delta_2=

Answer and worked solution

Official answer: 9

Triangle. P,Q=(h,±324−h2)P,Q=\left(h,\pm\frac{\sqrt3}2\sqrt{4-h^2}\right); the tangents meet on the x-axis at R(4h,0)R\left(\frac4h,0\right).

Area.
Δ(h)=12⋅34−h2⋅(4h−h)=3(4−h2)3/22h,\Delta(h)=\tfrac12\cdot\sqrt3\sqrt{4-h^2}\cdot\left(\tfrac4h-h\right)=\frac{\sqrt3(4-h^2)^{3/2}}{2h},
decreasing on [12,1]\left[\frac12,1\right].

Extremes. Δ1=Δ(12)=3(154)3/2=4558\Delta_1=\Delta\left(\tfrac12\right)=\sqrt3\left(\tfrac{15}4\right)^{3/2}=\dfrac{45\sqrt5}8, Δ2=Δ(1)=92\Delta_2=\Delta(1)=\dfrac92.

Value. 85⋅4558−8⋅92=45−36=9\dfrac8{\sqrt5}\cdot\dfrac{45\sqrt5}8-8\cdot\dfrac92=45-36=9.

Answer: 9

Topics:EllipseTangents and normals to conicsMaxima and minima
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