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The area of the region eclosed between the parabolas $\mathrm{y^{2}=2x-1}$ and $\mathrm{y^{2}=4x-3}$ is

Option: 1
$\mathrm{\frac{1}{3}}$

Option: 2
$\mathrm{\frac{1}{6}}$

Option: 3
$\mathrm{\frac{2}{3}}$

Option: 4
$\mathrm{\frac{3}{4}}$

Answers (1)

We have, $\mathrm{y^{2}=2x-1}$ ...........(1)

and $\mathrm{y^{2}=4x-3}$ .........(2)

We first find the point of inter section of the given parabolas by solving the eqn (1) and (2) simultaneously.

This gives, $\mathrm{2x-1=4x-3}$

$\Rightarrow \mathrm{2x=2\Rightarrow x=1}$

Now, put $\mathrm{x=1}$ in eqn (1), we get $\mathrm{y=\pm 1}$

Now the eqn(1) and (2) can be written as

$\mathrm{y^{2}=2\left ( x-\frac{1}{2} \right )}$

$\mathrm{y^{2}=4\left ( x-\frac{3}{4} \right )}$ respectively.

The graphs of these two curves are

Now, the eqn (1) and (2) can be written as

$\mathrm{x=\frac{y^{2}+1}{2}}\: \&\: \mathrm{x=\frac{y^{2}+3}{4}}$ respectively

So $\mathrm{A=2\int _{0}^{1}\left ( x_{2}-x_{1} \right )dy}$

( $\therefore$ Area is symmetrical about x-axis )

Where $\mathrm{x_{1}=\frac{y^{2}+1}{2}\: and \: x_{2}=\frac{y^{2}+3}{4}}$

$\mathrm{A=2\int _{0}^{1}\left ( \frac{y^{2}+3}{4}-\frac{y^{2}+1}{2} \right )dy\\$

$=\mathrm{\frac{1}{2}\left[y-\frac{y^{3}}{3}\right]_{0}^{1}=\frac{1}{2}\left[1-\frac{1}{3} \right ]}\\$

$=\mathrm{\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}}$ sq.units

Hence he correct answer is option 1.

Posted by

Kshitij

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