CUET PG 2024 — Mathematics PYQ

$y^2=4x$ vertex at (0,0) passes through $(4,\pm 4)$
$x^2-4y--$vertex at(0,0)

passes through

(4,4) $\&$ (4,-4)

$x^2=4(\underline{\sqrt{4x}}$ $(\because y^2=4x)$

$x^2=8\sqrt{x}$

$x^4=64x$

$x^3=64$

$x=4$

$\therefore y^2=16\Rightarrow y=\pm 4$

Intersection points are

(4,4),(4,-4)

Area= ${\int }_0^{4}2\int x-\frac{x^2}{4})$ $dx$

$={\left[\frac{4}{3}{x}^{3/2}-\frac{x^3}{12}\right]}_0^4$

$\left[\frac{4}{3}8-\frac{64}{12}\right]-0$

$=\frac{32}{3}-\frac{16}{3}$ $=\frac{16}{6}$