Question:
If the tangent at a point š on the parabola \(š¦^2=3š„\) is parallel to the line \(š„+2š¦=1\) and the tangents at the points š and š
on the ellipse \(\frac{š„^2}{ 4} +\frac{š¦^2}{ 1} =1\) are perpendicular to the line š„āš¦=2, then the area of the triangle PQR is :
If the tangent at a point š on the parabola \(š¦^2=3š„\) is parallel to the line \(š„+2š¦=1\) and the tangents at the points š and š
on the ellipse \(\frac{š„^2}{ 4} +\frac{š¦^2}{ 1} =1\) are perpendicular to the line š„āš¦=2, then the area of the triangle PQR is :
Updated On: Mar 12, 2026
- \(\frac{3}{2}\sqrt 5\)
- \(5\sqrt 3\)
- \(3\sqrt 5\)
- \(\frac{9}{\sqrt 5}\)
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The Correct Option is C
Solution and Explanation
(A)The tangent to the parabola \( y^2 = 3x \) has slope \( m \). The equation of the tangent is:
\[
y = mx + \frac{1}{m}.
\]
For parallelism to \( x + 2y = 1 \), \( m = -\frac{1}{2} \).
(B)Compute the coordinates of \( P \). Substituting \( y = -\frac{1}{2}x \) into \( y^2 = 3x \):
\[
\left(-\frac{1}{2}\right)^2x^2 = 3x \implies x = 4, y = -2.
\]
(C)For the ellipse, tangents perpendicular to \( x - y = 2 \) have slopes \( m = 1 \). Substituting into tangent conditions, find \( Q \) and \( R \).
(D)Use the vertices \( P, Q, R \) to find the area using the determinant formula:
\[
\text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|.
\]
Compute to get \( 3\sqrt{5} \).
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