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 (x2/4)+(y2/1)=1 are perpendicular to the line x-y=2, then the area of the triangle PQR is:

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 (x2/4)+(y2/1)=1 are perpendicular to the line x-y=2, then the area of the triangle PQR is: (A) (9/√5) (B) 5 √3 (C) (3/2) √5 (D) 3 √5

Tardigrade Admin · Accepted Answer

y2=3 x Tangent P ( x 1, y 1) is parallel to x +2 y =1 Then slope at P =-(1/2) 2 y ( dy / dx )=3 ⇒ ( dy / dx )=(3/2 y )=-(1/2) ⇒ y1=-3 Coordinates of P (3,-3) Similarly Q ((4/√3), (1/√5)), R (-(4/√5), (-1/√5)) Area of triangle PQR =(1/2)|3 -3 1 (4/√5) (1/√5) 1 -(4/√5) -(1/√5) 1| =(1/2)[3((2/√5))+3((8/√5))+0]=(30/2 √5)=3 √5

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:

$\frac{9}{5}$

$53$

$\frac{3}{2} 5$

$35$

Q. 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 4x2​+1y2​=1 are perpendicular to the line x−y=2, then the area of the triangle PQR is:

5​9​

53​

23​5​

35​

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:

$\frac{9}{5}$

$53$

$\frac{3}{2} 5$

$35$