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(D) The ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P = (h, y_0)$ and $Q = (h, -y_0)$. Since $P$ is on the ellipse,$\frac{h^2}{4}+\frac{y_0^2}{3}

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P = (h, y_0)$ and $Q = (h, -y_0)$. Since $P$ is on the ellipse,$\frac{h^2}{4}+\frac{y_0^2}{3}=1$,so $y_0 = \sqrt{3(1-\frac{h^2}{4})} = \frac{\sqrt{3}}{2}\sqrt{4-h^2}$.The tangent at $P(x_1, y_1)$ is $\frac{xx_1}{4}+\frac{yy_1}{3}=1$. For $P(h, y_0)$,the tangent is $\frac{xh}{4}+\frac{yy_0}{3}=1$. Setting $y=0$,we get $x = \frac{4}{h}$. Thus,$R = (\frac{4}{h}, 0)$.The area of $	riangle PQR$ is $\Delta(h) = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2y_0) 	imes (\frac{4}{h}-h) = y_0(\frac{4-h^2}{h}) = \frac{\sqrt{3}}{2}\sqrt{4-h^2} \cdot \frac{4-h^2}{h} = \frac{\sqrt{3}}{2} \frac{(4-h^2)^{3/2}}{h}$.Let $h = 2\cos 	heta$. Then $h \in [1/2, 1] \implies \cos 	heta \in [1/4, 1/2]$.$\Delta(	heta) = \frac{\sqrt{3}}{2} \frac{(4-4\cos^2 	heta)^{3/2}}{2\cos 	heta} = \frac{\sqrt{3}}{2} \frac{8\sin^3 	heta}{2\cos 	heta} = 2\sqrt{3} \frac{\sin^3 	heta}{\cos 	heta}$.Since $\frac{d\Delta}{d	heta} > 0$,$\Delta$ is increasing with $	heta$. As $\cos 	heta$ decreases from $1/2$ to $1/4$,$	heta$ increases,so $\Delta$ increases.$\Delta_2 = \Delta(h=1) = \frac{\sqrt{3}}{2} \frac{(4-1)^{3/2}}{1} = \frac{3\sqrt{3}\sqrt{3}}{2} = \frac{9}{2} = 4.5$.$\Delta_1 = \Delta(h=1/2) = \frac{\sqrt{3}}{2} \frac{(4-1/4)^{3/2}}{1/2} = \sqrt{3} (15/4)^{3/2} = \sqrt{3} \frac{15\sqrt{15}}{8} = \frac{45\sqrt{5}}{8}$.Finally,$\frac{8}{\sqrt{5}} \Delta_1 - 8 \Delta_2 = \frac{8}{\sqrt{5}} \cdot \frac{45\sqrt{5}}{8} - 8 \cdot \frac{9}{2} = 45 - 36 = 9$.

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