If the area bounded by the curves y=ax^2 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given curves are $y=ax^2$ $(i)$ and $x=ay^2$ $(ii)$.Substituting $y$ from $(ii)$ into $(i)$,we get $x = a(ax^2)^2 = a^3x^4$,which implies $x(a^3x^

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Given curves are $y=ax^2$ $(i)$ and $x=ay^2$ $(ii)$.Substituting $y$ from $(ii)$ into $(i)$,we get $x = a(ax^2)^2 = a^3x^4$,which implies $x(a^3x^3 - 1) = 0$.Thus,$x=0$ or $x=\frac{1}{a}$.For $x=0$,$y=0$,and for $x=\frac{1}{a}$,$y=\frac{1}{a}$.The area bounded by the curves is given by $\int_0^{1/a} (\sqrt{x/a} - ax^2) dx = 3$.$\int_0^{1/a} \frac{1}{\sqrt{a}} x^{1/2} dx - \int_0^{1/a} ax^2 dx = 3$.$\frac{1}{\sqrt{a}} [\frac{2}{3} x^{3/2}]_0^{1/a} - a [\frac{x^3}{3}]_0^{1/a} = 3$.$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} \cdot (\frac{1}{a})^{3/2} - \frac{a}{3} \cdot (\frac{1}{a})^3 = 3$.$\frac{2}{3a^2} - \frac{1}{3a^2} = 3$.$\frac{1}{3a^2} = 3 \Rightarrow 9a^2 = 1 \Rightarrow a^2 = \frac{1}{9}$.Since $a>0$,we have $a = \frac{1}{3}$.

If the area bounded by the curves $y=ax^2$ and $x=ay^2$ $(a>0)$ is $3$ sq units,then the value of $a$ is

Similar Questions

The area of the region described by $A = \{(x,y) : x^2 + y^2 \le 1 \text{ and } y^2 \le 1-x \}$ is

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

The area of the region bounded by the curve $y = \max \{| x |, x | x - 2 |\}$,the $x$-axis,and the lines $x = -2$ and $x = 4$ is equal to . . . . . . .

The area between the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$ and the straight line $y = 2$ is:

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module