Ncert Solutions Maths class 12th

The given equation of the curve is

$\begin{array}{l}{y}^2=4x\\ \Rightarrow y=\pm \surd 3\end{array}$

$\begin{array}{l}\mathrm{\to \; y}=+\surd 2x\end{array}$ in Ist quadrant

So, area of curve enclosed by $y^2=4x$

And $x=3=2\times \; area$ area (AOCA)

Given curve is $x^2=4y$ and the equation of line is $x=4y-2$

The point of intersection of the curve and the line can be determine as follows.

Put, $x=4y-2\Rightarrow x+2=4y\Rightarrow y\frac{x+2}{4}$

In $x^2=4y$ to determine value of x

i.e, $x^2=4\times \frac{\left(x+2\right)}{4}=x+2$

$\begin{array}{l}\Rightarrow x^2-x-2=0\Rightarrow x^2+x-2x-2=0\\ \Rightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Rightarrow \left(x+1\right)\left(x-2\right)=0\end{array}$

$\Rightarrow x=2$ and $x=-1$

$x=2$ , we have ${\left(2\right)}^2=4y\Rightarrow 4-4y\Rightarrow y=1$

And at $x=-1$ we have ${\left(-1\right)}^2=4y\Rightarrow y=\frac{1}{4}$

So, the coordinates A and B are (2,1)

Given that equation of

curve $y=x^2$

line $y=\left|x\right|=\left\{x,ifx\ge 0\&-x,ifx\angle 0\right\}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y=x^2$

$\Rightarrow x=x^2$ for Ist quadrant and

$\Rightarrow -x=x^2$ for IInd t quadrant

So, $\Rightarrow x^2-x=0$ and $x^2+x=0$

$\Rightarrow x\left(x-1\right)=0$ and $x\left(x+1\right)=0$

$\Rightarrow x=0,1$ $x=0,-1$

$x=0,y=0$

$x=1,y=1$ i.e, A has coordinate (1,1)

$x=-1,y=1$ i.e, B has coordinate (1,1)

Now, area o

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$\Rightarrow 4a^3=4^3$ (Squaring both sides)

$\Rightarrow a^3=4^2$

$\Rightarrow a=4^{\frac{2}{3}}$ (Taking cube on both sides)

Kindly go through the solution

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Kindly go through the solution