For a > 0, let the curves C _ 1 : y ^ 2 = ax and C _ 2 : x ^ 2 = ay intersect at origin O and a point P . Let the line x = b (0 < b < a) intersect the chord OP and the x -axis at points Q and R, respectively. If the

Correct answer: B. x ^ 6 - 12 x ^ 3 + 4 = 0 ∫ _ 0 ^ b ( ax - x ^ 2 a )dx = a ^ 2 6 ⇒ (2)/(3) a b(3)/(2) - b ^ 3 3a = a ^ 2 6 . . . (1) Also area of OQR = (1)/(2) (1)/(2) b ^ 2 = (1)/(2) ⇒ b = 1 Put in (1) ⇒ 4a a - 2 = a ^ 3 ⇒ a ^ 6 + 4 a ^ 3 + 4 = 16 a ^ 3 ⇒ a ^ 6 - 12 a ^ 3 + 4 = 0

JEE Main 20208 JanuaryMorning Shift

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

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

For a > 0, let the curves C _ 1 : y ^ 2 = ax and C _ 2 : x ^ 2 = ay...

For

a > 0,

let the curves

{C}_{1}: {y}^{2} = ax and

{C}_{2}: {x}^{2} = ay

intersect at origin

and a point

P .

Let the line

x = b

\left(0 < b < a\right)

intersect the chord

OP

and the

x

-axis at points

Q

and

R,

respectively. If the line

x = b

bisects the area bounded by the curves,

{C}_{1}

and

{C}_{2},

and the area of

∆ OQR = \dfrac{1}{2},

then ‘

a

’ satisfies the equation:

Options

A ${x}^{6} - 6{x}^{3} + 4 = 0$
B ${x}^{6} - 12{x}^{3} + 4 = 0$
C ${x}^{6} + 6{x}^{3} - 4 = 0$
D ${x}^{6} - 12{x}^{3} - 4 = 0$

Correct answer

B. ${x}^{6} - 12{x}^{3} + 4 = 0$

Step-by-step solution

\mathop{∫}_{0}^{b}\left(\sqrt{ax} - \dfrac{{x}^{2}}{a}\right)dx = \dfrac{{a}^{2}}{6} $$ ⇒ \dfrac{2}{3}\sqrt{a}b\dfrac{3}{2} - \dfrac{{b}^{3}}{3a} = \dfrac{{a}^{2}}{6} . . . \left(1\right) $Also area of$ \delta OQR = \dfrac{1}{2} $$ \dfrac{1}{2}{b}^{2} = \dfrac{1}{2} ⇒ b = 1 $Put in \left(1\right)$

$⇒ 4a\sqrt{a} - 2 = {a}^{3}$

$⇒ {a}^{6} + 4{a}^{3} + 4 = 16{a}^{3}$

$⇒ {a}^{6} - 12{a}^{3} + 4 = 0$

Practice Parabola on Quantrex Academy →

More from Parabola