Question

If the tangent at a point $P$ on the parabola $y ^2=3 x$ is parallel to the line $x+2 y=1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are perpendicular to the line $x-y=2$, then the area of the triangle $PQR$ is:

• A. $\frac{9}{\sqrt{5}}$
• B. $5 \sqrt{3}$
• C. $\frac{3}{2} \sqrt{5}$
• D. $3 \sqrt{5}$

Answer 1

Answer: (D)

$y^2=3 x$
Tangent $P \left( x _1, y _1\right)$ is parallel to $x +2 y =1$
Then slope at $P =-\frac{1}{2}$
$2 y \frac{ dy }{ dx }=3$
$ \Rightarrow \frac{ dy }{ dx }=\frac{3}{2 y }=-\frac{1}{2} $
$ \Rightarrow y_1=-3$
Coordinates of $P (3,-3)$
Similarly $Q \left(\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{5}}\right), R \left(-\frac{4}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)$
Area of $\triangle PQR$
$=\frac{1}{2}\begin{vmatrix}3 & -3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ -\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1\end{vmatrix}$
$=\frac{1}{2}\left[3\left(\frac{2}{\sqrt{5}}\right)+3\left(\frac{8}{\sqrt{5}}\right)+0\right]=\frac{30}{2 \sqrt{5}}=3 \sqrt{5}$