Single Choice

A vertical line passing through the point $$(\mathrm{h}, 0)$$ intersects the ellipse $$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. Let the tangents to the ellipse at $$\mathrm{P}$$ and $$\mathrm{Q}$$ meet at the point $$\mathrm{R}$$. If $$\Delta(\mathrm{h})=$$ area of the triangle $$\mathrm{P}\mathrm{Q}\mathrm{R},\ \Delta_{1}= \underset{1/2\leq h\leq 1}{max} \Delta(\mathrm{h})$$ and $$\displaystyle \Delta_{2}=\min_{1/2\leq h\leq 1}\Delta(\mathrm{h})$$ , then $$\displaystyle \frac{8}{\sqrt{5}}\Delta_{1}-8\Delta_{2}=$$

A$$3$$
B$$6$$
C$$9$$
Correct Answer
D$$12$$

Solution

Solution:
$$\cfrac{x^2}{4}+\cfrac{y^2}{3}=1$$

$$y=\cfrac{\sqrt3}{2}\sqrt{4-h^2}$$ at $$x=h$$

Let $$R=(x_1,0)$$

$$PQ$$ is a chord of contact, so $$\cfrac{xx_1}{4}=1\Longrightarrow x=\cfrac{4}{x_1}$$

which is equation of $$PQ$$, $$x=h$$

so, $$\cfrac{4}{x_1}=h\Longrightarrow x_1=\cfrac{4}{h}$$

$$\triangle(h)=ar(\triangle PQR)=\cfrac12\times PQ\times RT$$

$$\cfrac12\times \cfrac{2\sqrt3}{2}\sqrt{4-h^2}\times (x_1-h)$$

$$=\cfrac{\sqrt3}{2h}(4-h^2)^{3/2}$$

$$\triangle '(h)=\cfrac{-\sqrt3(4+2h^2)}{2h^2}\sqrt{4-h^2}$$ which is always decreasing.

so $$\triangle_1=$$ maximum of $$\triangle(h)=\cfrac{45\sqrt5}{8}$$ at $$h=\cfrac12$$

$$\triangle_2=$$ minimum of $$\triangle(h)=\cfrac{9}{2}$$ at $$h=1$$

so $$\cfrac{8}{\sqrt5}\triangle_1-8\triangle_2=\cfrac{8}{\sqrt5}\times\cfrac{45\sqrt5}{8}-8\times\cfrac92=45-36=9$$

Hence, C is the correct option.

SIMILAR QUESTIONS

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The area (in sq. units) of the quadrilateral formed by the tangent at the end points of the latera recta to the ellipse $$\displaystyle \dfrac {x^2}{9}+\dfrac {y^2}{5}=1$$, is

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The minimum area of the triangle formed by any tangent to the ellipse $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with the coordinate axes is

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