JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
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 line $x=b$ bisects the area bounded by the curves, $C_1$ and $C_2$, and the area of $\Delta OQR=\frac12$, then `a' satisfies the equation :
Application of Integrals
Solution
$P=(a,a)$, chord $OP$: $y=x$. $Q=(b,b)$, $R=(b,0)$, so $\frac{b^2}2=\frac12\Rightarrow b=1$. Total area between curves $=\int_0^a(\sqrt{ax}-x^2/a)dx=\frac{a^2}3$. Half: $\int_0^1(\sqrt{ax}-x^2/a)dx=\frac23\sqrt a-\frac1{3a}=\frac{a^2}6$. Multiply by $6a$: $4a^{3/2}-2=a^3\Rightarrow a^3+2=4a^{3/2}$. Squaring: $a^6+4a^3+4=16a^3\Rightarrow a^6-12a^3+4=0$. Option (B).