If the area bounded by the curve y 2 = 4 a x and y = m x is a 2 3 , then the value of m is:

Question

If the area bounded by the curve $y^{2} = 4 a x$ and $y = m x$ is $\frac{a^{2}}{3}$ , then the value of $m$ is:

Smart Achievers · Accepted Answer

(1)

Ans. 2

Explanation:

Given, $y = m x$ $\dots (i)$

$y = m x$ $\dots (i i)$

Substitute (ii) in (i):

${(m x)}^{2} = 4 a x$

which gives $m^{2} x^{2} - 4 a x = 0$

$x [m^{2} x - 4 a] = 0$

i.e., $x = 0, x = \frac{4 a}{m^{2}}$

From (ii), equation $x = 0$ , $y = 0$

and $x = \frac{4 a}{m^{2}}$

$\Rightarrow$ $y = m \times (\frac{4 a}{m^{2}})$

$y = \frac{4 a}{m}$

$∴$ $Area = \int_{0}^{\frac{4 a}{m^{2}}} y d x - \int_{0}^{\frac{4 a}{m^{2}}} m x d x$

$= \int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x$

$= 2 \sqrt{a} {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{\frac{4 a}{m^{2}}} - m {[\frac{x^{2}}{2}]}_{0}^{\frac{4 a}{m^{2}}}$

$= \frac{2 \sqrt{a} \times 2}{3} [{(\frac{4 a}{m^{2}})}^{3 / 2}] - \frac{m}{2} {[\frac{4 a}{m^{2}}]}^{2} - (0) - (0)$

$= \frac{4 \sqrt{a}}{3} [\frac{8 a^{3 / 2}}{m^{3}}] - \frac{16 a^{2}}{2 m^{3}}$

$= \frac{32 a^{2}}{3 m^{3}} - \frac{16 a^{2}}{2 m^{3}}$

Now it is given

$\frac{32 a^{2}}{3 m^{3}} - \frac{16 a^{2}}{2 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{64 a^{2} - 48 a^{2}}{6 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{16 a^{2}}{6 m^{3}} = \frac{a^{2}}{3}$

$\Rightarrow$ $\frac{16}{6} \times 3 = m^{3}$

$\Rightarrow$ $m^{3} = 8$

$∴$ $m = 2$

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