Question

Assertion(A): The area bounded by $y^2=4x$ and $x^2=4y$ is $\frac{16}{3}$ sq. units.

Reason(R): The area bounded by $y^2=4ax$ and $x^2=4ay$ is $\frac{16a^2}{3}$ sq. units

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true but R is not the correct explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A.

Given $y^2=4ax$ and $x^2=4ay$

Since both curves intersect,

$\therefore x=\frac{y^2}{4a}$

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

$\Rightarrow y^4=64a^{3}y$

$\Rightarrow y[y^3-(4a{)}^3]=0$

$\Rightarrow y=0,4a.$

When $y=0,x=0$ and when $y=4a,x=4a,$

Area $A={\int }_0^{4a}\left(\sqrt{(}4ax\right)-\frac{x^2}{4a})dx$

on solving we will get

$A=\frac{16}{3a^2}$

Now if we put $a=1$,

we will get $A=\frac{16}{3}$