If the tangent at a point P on the parabola y2=3x is parallel to the line x+2y=1 and the tangents at the points Q and R on the ellipse x24+y21=1 are perpendicular to the line x−y=2 , then the area of the triangle is PQR is:

y2=3xTangent P(x1,y1) is parallel to x+2y=1Then slope at P=−122ydydx=3⇒dydx=32y=−12⇒y1=−3Coordinates of P(3,−3)Similarly Q(43,15),R(−45,−15)B(4+22,1+22)x−(2+22)1=y−(3+22)−1=2∴ area of trapezium:12(4+42)2=4(1+2)

If the tangent at a point P on the parabola y2=3x is parallel to the line x+2y=1 and the tangents at the points Q and R on the ellipse x24+y21=1 are perpendicular to the line x−y=2 , then the area of the triangle is PQR is:

y2=3xTangent P(x1,y1) is parallel to x+2y=1Then slope at P=−122ydydx=3⇒dydx=32y=−12⇒y1=−3Coordinates of P(3,−3)Similarly Q(43,15),R(−45,−15)B(4+22,1+22)x−(2+22)1=y−(3+22)−1=2∴ area of trapezium:12(4+42)2=4(1+2)

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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 is $P Q R$ is:

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

$5 \sqrt{3}$

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

$3 \sqrt{5}$

answer is D.

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Detailed Solution

$y^{2} = 3 x$
Tangent $P (x_{1}, y_{1})$ is parallel to $x + 2 y = 1$
Then slope at $P = - \frac{1}{2}$
$2 y \frac{d y}{d x} = 3$
$\Rightarrow \frac{d y}{d x} = \frac{3}{2 y} = - \frac{1}{2}$
$\Rightarrow y_{1} = - 3$
Coordinates of $P (3, - 3)$
Similarly $Q (\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{5}}), R (- \frac{4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$ $B (4 + 2 \sqrt{2}, 1 + 2 \sqrt{2})$
$\frac{x - (2 + 2 \sqrt{2})}{1} = \frac{y - (3 + 2 \sqrt{2})}{- 1} = 2$
$∴$ area of trapezium:
$\frac{1}{2} (4 + 4 \sqrt{2}) 2 = 4 (1 + \sqrt{2})$