The area enclosed between the curves $y = {ax}^{2}$ and $x = {ay}^{2}(a 0)$ is 1 sq unit. Then value of $a$ is

Question

The area enclosed between the curves $y = {ax}^{2}$ and $x = {ay}^{2}(a > 0)$ is 1 sq unit. Then value of $a$ is

Dr. Abubacker · Accepted Answer

(a)
The points of intersection of given curves are (0,0) and
$\left( \frac{1}{a},\frac{1}{a} ight)$
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$	herefore\ \ Required\ area\ OABCO = area\ of\ OCBDO - area\ of\ OABDO$
$\Longrightarrow \ \ \int_{0}^{1/a}{\left( \sqrt{\frac{x}{a}} - {ax}^{2} ight)dx = 1}\ \ \ \lbrack givenbrack$
$\Longrightarrow \ \ \left( \frac{1}{\sqrt{a}} \bullet \frac{x^{3/2}}{3/2} - \frac{{ax}^{3}}{3} ight)_{0}^{1/a} = 1$
$\Longrightarrow \ \frac{2}{{3a}^{2}} - \frac{1}{{3a}^{2}} = 1$
$\Longrightarrow \ \ a^{2} = \frac{1}{3}\  \Longrightarrow a = \frac{1}{\sqrt{3}}\ \ \ \ \ \ \ \ \lbrack as\ a > 0brack$

Q. The area enclosed between the curves $y = {ax}^{2}$ and $x = {ay}^{2}(a > 0)$ is 1 sq unit. Then value of $a$ is

Explanation

Reviewed by Dr. Abubacker

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