Let a line y = mx (m > 0), intersect the parabola, y ^ 2 = x , at a point P, other than the origin. Let the tangent to it a P , meet the x -axis at the point Q . If area ( OPQ) = 4 square unit, then m is equal to

Correct answer: 0 Let the equation of tangent at P is 2ty = x + t ^ 2 So, Q(- t ^ 2 , 0) As given, area ( OPQ) = 4 square unit, so (1)/(2)| cases 0 & 0 & 1 t ^ 2 & t & 1 - t ^ 2 & 0 & 1 cases | = ± 4 |t| ^ 3 = ± 8 t = 2(t > 0) So, m = t t ^ 2 = (1)/(2).

JEE Main 20208 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2020 · 8 January · Evening Shift · Mathematics · Parabola

Let a line y = mx (m > 0), intersect the parabola, y ^ 2 = x , at a...

Let a line

y = mx \left(m > 0\right), intersect

the parabola,{y}^{2} = x

, at a point

other than the origin. Let the tangent to it a

, meet the

-axis at the point

Q .

If area \left(\delta OPQ\right) = 4

square unit, then

m

is equal to Let a line y = mx (m > 0), intersect the parabola, y ^ 2 = x , at a point P, other than the origin. Let the tangent to

Correct answer

Step-by-step solution

Let a line y = mx (m > 0), intersect the parabola, y ^ 2 = x , at a point P, other than the origin. Let the tangent to
Let the equation of tangent at $P$ is $2ty = x + {t}^{2}$
So, $Q\left(- {t}^{2}, 0\right)$

As given, area \left(\delta OPQ\right) = 4 $square unit, so $ \dfrac{1}{2}\left| $\begin{cases}0 & 0 & 1 \\ {t}^{2} & t & 1 \\ - {t}^{2} & 0 & 1\end{cases}$ \right| = \pm 4 $ $ {\left|t\right|}^{3} = \pm 8 $ $ t = 2\left(t > 0\right) $So,$ m = \dfrac{t}{{t}^{2}} = \dfrac{1}{2}.

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