The area between the parabola is $y^{2} = 2 a x a n d x^{2} = 2 a y$

\frac{2 a^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 a^{2}}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 a^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8 a^{2}}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{4 a^{2}}{3}$

The parabola intersect at 0 (0, 0) and A (2a, 2a)

Required area $= \int_{0}^{2 a} \int_{x^{2} / 2 a}^{\sqrt{2 a x}} d y d x$

$= \int_{0}^{2 a} (\sqrt{2 a x} - \frac{x^{2}}{2 a}) d x$

$= {∣ ∣ ∣ \sqrt{2 a} \frac{2}{3} x^{3 / 2} - \frac{1}{2 a} \frac{x^{3}}{3} ∣ ∣ ∣}_{0}^{3 / 2}$

$= \sqrt{2 a} \frac{2}{3} (2 a)^{3 / 2} - \frac{1}{2 a} \frac{(2 a)^{3}}{3}$

$= \frac{8 a^{2}}{3} - \frac{4 a^{2}}{3} = \frac{4 a^{2}}{3}$

Suggest Corrections

Similar questions

ABCD is a square, the length of whose side is a. Taking AB and AD as the coordinate axes, the equation of the

circle passing through the vertices of the square, is

x^{2} - 2 a x + 2 a y = 0

x^{2} + y^{2} + 2 a x + 2 b y + c = 0

x^{2} + y^{2} + 2 b x + 2 a y + c = 0

x^{2} - 2 a x + 2 a y = 0 .

The equation of the locus of a point which moves so as to be at equal distances from the point (a, 0) and the $y^{-}$ axis is

Join BYJU'S Learning Program