Answer
Let $P(a,2a^2)$
Let A=The area between curve $y=2x^2$ and $y=x^2$
$A=\int_0^a2x^2-x^2dx=\frac13x^3|_0^a=\frac{a^3}3$
B=The area between curve $y=2x^2$ and C
Because $y=2x^2,x=\sqrt{\frac y2}$
Consider the curve C as $x=C(y)$
$B=\int_0^{2a^2}\sqrt{\frac y2}-C(y)dy$
$B=[\frac43(\frac y2)^{\frac32}]_0^{2a^2}-\int_0^{2a^2}C(y)dy$
$B=\frac43a^3\int_0^{2a^2}C(y)dy$
Because $A=B$
$\int_0^{2a^2}C(y)dy=a^3$
$\left(\int_0^{2a^2}C(y)dy\right)'=(a^3)'$
$4aC(2a^2)=3a^2$
let $y=2a^2,a=\sqrt{\frac y2}$
So $C(y)=\frac34\sqrt{\frac y2}$
As $C(y)=x$
$\frac34\sqrt{\frac y2}=x$
$y=\frac{32}9x^2$
Work Step by Step
Let $P(a,2a^2)$
Let A=The area between curve $y=2x^2$ and $y=x^2$
$A=\int_0^a2x^2-x^2dx=\frac13x^3|_0^a=\frac{a^3}3$
B=The area between curve $y=2x^2$ and C
Because $y=2x^2,x=\sqrt{\frac y2}$
Consider the curve C as $x=C(y)$
$B=\int_0^{2a^2}\sqrt{\frac y2}-C(y)dy$
$B=[\frac43(\frac y2)^{\frac32}]_0^{2a^2}-\int_0^{2a^2}C(y)dy$
$B=\frac43a^3\int_0^{2a^2}C(y)dy$
Because $A=B$
$\int_0^{2a^2}C(y)dy=a^3$
$\left(\int_0^{2a^2}C(y)dy\right)'=(a^3)'$
$4aC(2a^2)=3a^2$
let $y=2a^2,a=\sqrt{\frac y2}$
So $C(y)=\frac34\sqrt{\frac y2}$
As $C(y)=x$
$\frac34\sqrt{\frac y2}=x$
$y=\frac{32}9x^2$
