A vertical line passing through the point ( h , 0 ) intersects the ellipse x 2 4 + y 2 3 = 1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If ( h ) is the area of the triangle PQR, 1 = max 1 / 2 h 1 ( h ) an

Question

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If $Δ (h)$ is the area of the triangle PQR, $Δ_{1} = \max_{1 / 2 \leq h \leq 1} Δ (h) and Δ_{2} = \min_{1 / 2 \leq h \leq 1} Δ (h),$ then $\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} =$ [2013]

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(9)

$Vertical line x = h, meets the ellipse \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1 at$

$P (h, \frac{\sqrt{3}}{2} \sqrt{4 - h^{2}}) and Q (h, \frac{- \sqrt{3}}{2} \sqrt{4 - h^{2}})$

$By symmetry, tangents at P and Q will meet each other at x-axis.$

$Tangent at P is \frac{x h}{4} + \frac{y \sqrt{3}}{6} \sqrt{4 - h^{2}} = 1,$ $which meets x-axis at R (\frac{4}{h}, 0)$

$area (∆ P Q R) = \frac{1}{2} \times \sqrt{3} \sqrt{4 - h^{2}} \times (\frac{4}{h} - h)$

$Let Δ (h) = \frac{\sqrt{3} {(4 - h^{2})}^{3 / 2}}{2 h}$

$\Rightarrow \frac{d Δ}{d h} = - \sqrt{3} [\frac{\sqrt{4 - h^{2}} (h^{2} + 2)}{h^{2}}] < 0$

$∴$ $Δ (h) is a decreasing function.$

$∴$ $\frac{1}{2} \leq h \leq 1 \Rightarrow Δ_{\max} = Δ (\frac{1}{2}) and Δ_{\min} = Δ (1)$

$∴$ $Δ_{1} = Δ_{\max} = \frac{\sqrt{3}}{2} \cdot \frac{{(4 - \frac{1}{4})}^{3 / 2}}{\frac{1}{2}} = \frac{45}{8} \sqrt{5}$ $and Δ_{2} = Δ_{\min} = \frac{\sqrt{3} \cdot 3 \sqrt{3}}{2 \cdot 1} = \frac{9}{2}$

$∴$ $\frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} = 45 - 36 = 9$

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Q. A vertical line passing through the point (h,0) intersects the ellipse x24+y23=1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ(h) is the area of the triangle PQR, Δ1=max1/2≤h≤1Δ(h) and Δ2=min1/2≤h≤1Δ(h), then 85Δ1-8Δ2= [2013]

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