Ellipse Question 3
Question 3 - 29 January - Shift 2
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:
(1) $\frac{9}{\sqrt{5}}$
(2) $5 \sqrt{3}$
(3) $\frac{3}{2} \sqrt{5}$
(4) $3 \sqrt{5}$
Show Answer
Answer: (4)
Solution:
Formula: Equation of Tangent( Cartesian form ), Condition for Parallel Lines, Area of Triangle
$ y^2=3 x $
Tangent $P\left(x_1, y_1\right)$ is parallel to $x+2 y=1$
Then slope at $P=-\frac{1}{2}$
$ \begin{aligned} & 2 y \frac{dy}{dx}=3 \\ & \Rightarrow \frac{dy}{dx}=\frac{3}{2 y}=-\frac{1}{2} \\ & \Rightarrow y_1=-3 \end{aligned} $
Coordinates of $P(3,-3)$
Similarly $Q\left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right), R\left(-\frac{4}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)$
Area of $\triangle PQR$
$ \begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 3 & -3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ -\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1 \end{array}\right| \\ & =\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} \end{aligned} $
$\therefore$ The area of the triangle $PQR$ is $3 \sqrt{5}$
