For a0, let the curves C_1 : y^2=ax and C_2 : x^2=ay intersect at the origin O and a poin…

Question

For $a>0$, let the curves $C_1 : y^2=ax$ and $C_2 : x^2=ay$ intersect at the 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 $C_1$, $C_2$, and the area of $	riangle OQR = \dfrac12$, then $a$ satisfies the equation:

Accepted Answer

The curves meet at $O(0,0)$ and $P(a,a)$, so line $OP$ is $y=x$, and $Q=(b,b)$.

From $\triangle OQR$ area $=\dfrac12 b^2 = \dfrac12 \Rightarrow b=1$.

The bisection condition (area from $0$ to $b$ equals half the total area from $0$ to $a$):

$$\int_0^{b}\left(\sqrt{ax}-\frac{x^2}{a}\right)dx = \frac12\int_0^a\left(\sqrt{ax}-\frac{x^2}{a}\right)dx$$

With $b=1$, this simplifies (as in the source) to:

$$4a\sqrt a - 2 = a^3 \implies a^6+4a^3+4 = 16a^3 \implies a^6-12a^3+4=0$$

Solutions · 1