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 x 2 4 + y 2 1 = 1 are perpendicular to the line x - y = 2 , then the area of the triangle PQR is

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Solution

Q.
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 [2023]

1 $3 \sqrt{5}$

2 $\frac{9}{\sqrt{5}}$

3 $\frac{3}{2} \sqrt{5}$

4 $5 \sqrt{3}$

Ans.
(1)

If tangent at a point $P$ on $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and tangent at point $Q$ and $R$ on ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ .

Firstly we have equation of parabola $y^{2} = 3 x \dots (i)$

Tangent at $P (x_{1}, y_{1})$ is parallel to $x + 2 y = 1$

   $2 y = 1 - x$

   $y = - \frac{x}{2} + \frac{1}{2}$

{On comparing it with $y = m x + C$ }

Then slope, $m$ at $P = - \frac{1}{2} \dots (i i)$

On differentiating equation (i) with respect to $' x'$

$2 y \cdot \frac{d y}{d x} = 3 \Rightarrow \frac{d y}{d x} = \frac{3}{2 y} = - \frac{1}{2} (from (ii))$

$\Rightarrow y_{1} = - 3$

Co-ordinates of $P$ are $(3, - 3)$ .

Similarly, $Q (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}), R (\frac{- 4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$

So, we have three points P, Q and R by which area of $∆ P Q R$

$= \frac{1}{2}$ $| \begin{matrix} 3 & - 3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ \frac{- 4}{\sqrt{5}} & \frac{- 1}{\sqrt{5}} & 1 \end{matrix} |$    ${∴ Area of ∆ is given as = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |}$

$= \frac{30}{2 \sqrt{5}} = 3 \sqrt{5}$

Hence, the area of $∆ P Q R = 3 \sqrt{5}$ .

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