The area bounded by the parabola $y^{2} = 4 a x$ and the line $x = a$ and $x = 4 a$ is

\frac{35 a^{2}}{3}

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

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\frac{7 a^{2}}{3}

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\frac{56 a^{2}}{3}

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Solution

The correct option is B $\frac{56 a^{2}}{3}$
Required area $=$ area of curve PSMNRQP
$= 2$ area of curve PSRQP
$= 2 \int_{a}^{4 a} \sqrt{4 a x} d x$
$= 4 \sqrt{a} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}_{a}^{4 a} = \frac{8}{3} \sqrt{a} [(4 a)^{\frac{3}{2}} - a^{\frac{3}{2}}]$
$= \frac{8}{3} \sqrt{a} [8 a^{\frac{3}{2}} - a^{\frac{3}{2}}) = \frac{56}{3} a^{2}$

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Similar questions

y^{2} = 4 a (x + a)

y^{2} = - 4 a (x - a)

y = x^{2} + 1

x + y = 3

List-I	List-II
1. Area bounded by $y^{2} = 4 a x$ and $x^{2} = 4 a y$ is	a. $\frac{8 a^{2}}{3}$
2. Area bounded by $y^{2} = 4 a x$ and its latusrectum is	b. $\frac{a^{2}}{6}$
3. Area bounded by $x = 2 c o s θ, y = 2 s i n θ$ is	c. $\frac{16 a^{2}}{3}$
4. Area bounded by $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is	d. $2 π$

y^{2} = 4 a x

y^{2} = 16 x

x = 3

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