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 $\Delta \left(h\right)=$ area of the triangle $PQR,{\Delta }_1=\underset{1/2\le h\le 1}{max}\Delta \left(h\right)$ and ${\Delta }_2=\underset{1/2\le h\le 1}{min}\Delta \left(h\right)$ , then $\frac{8}{\sqrt{5}}{\Delta }_1-8{\Delta }_2=$

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is C $9$

Solution:

$\frac{x^2}{4}+\frac{y^2}{3}=1$

$y=\frac{\sqrt{3}}{2}\sqrt{4-h^2}$ at $x=h$

Let $R=(x_1,0)$

$PQ$ is a chord of contact, so $\frac{x{x}_1}{4}=1⟹x=\frac{4}{x_1}$

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

so, $\frac{4}{x_1}=h⟹x_1=\frac{4}{h}$

$△\left(h\right)=ar\left(△PQR\right)=\frac{1}{2}\times PQ\times RT$

$\frac{1}{2}\times \frac{2\sqrt{3}}{2}\sqrt{4-h^2}\times (x_1-h)$

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

${△}^{\prime }\left(h\right)=\frac{-\sqrt{3}(4+2h^2)}{2h^2}\sqrt{4-h^2}$ which is always decreasing.

so ${△}_1=$ maximum of $△\left(h\right)=\frac{45\sqrt{5}}{8}$ at $h=\frac{1}{2}$

${△}_2=$ minimum of $△\left(h\right)=\frac{9}{2}$ at $h=1$

so $\frac{8}{\sqrt{5}}{△}_1-8{△}_2=\frac{8}{\sqrt{5}}\times \frac{45\sqrt{5}}{8}-8\times \frac{9}{2}=45-36=9$

Hence, C is the correct option.