For a0, let the curves C_1: y^2=a x and C | vedclass

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

Question

(A) The curves are $C_{1}: y^{2}=ax$ and $C_{2}: x^{2}=ay.$ Solving them,we get $x^{4}/a^{2} = ax \Rightarrow x(x^{3}-a^{3})=0.$ Thus,the intersection

Vedclass · Accepted Answer

$a^{6}-12 a^{3}+4=0$

Vedclass · Answer

(A) The curves are $C_{1}: y^{2}=ax$ and $C_{2}: x^{2}=ay.$ Solving them,we get $x^{4}/a^{2} = ax \Rightarrow x(x^{3}-a^{3})=0.$ Thus,the intersection points are $O(0,0)$ and $P(a,a).$
The chord $OP$ has the equation $y=x.$ The line $x=b$ intersects $OP$ at $Q(b,b)$ and the $x$-axis at $R(b,0).$
The area of $\Delta OQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes b 	imes b = \frac{b^{2}}{2}.$ Given $	ext{Area} = \frac{1}{2},$ we have $b^{2}=1 \Rightarrow b=1.$
The total area bounded by the curves is $\int_{0}^{a} (\sqrt{ax} - x^{2}/a) dx = [\frac{2}{3}\sqrt{a}x^{3/2} - x^{3}/(3a)]_{0}^{a} = \frac{2}{3}a^{2} - \frac{1}{3}a^{2} = \frac{a^{2}}{3}.$
The line $x=b$ bisects this area,so $\int_{0}^{b} (\sqrt{ax} - x^{2}/a) dx = \frac{1}{2} 	imes \frac{a^{2}}{3} = \frac{a^{2}}{6}.$
Substituting $b=1$: $\frac{2}{3}\sqrt{a} - \frac{1}{3a} = \frac{a^{2}}{6}.$
Multiplying by $6a$: $4a^{3/2} - 2 = a^{3} \Rightarrow a^{3} + 2 = 4a^{3/2}.$
Squaring both sides: $(a^{3}+2)^{2} = 16a^{3} \Rightarrow a^{6} + 4a^{3} + 4 = 16a^{3} \Rightarrow a^{6} - 12a^{3} + 4 = 0.$

Similar Questions

The area bounded by the curves $|x| + |y| \geq 1$ and $x^2 + y^2 \leq 1$ is

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

The area bounded by the curve $y=x^2+3$,$y=x$,$x=3$ and the $y$-axis is:

The area (in sq. units) of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2\}$ is

The area of the region bounded by $y^2=x$ and $y=|x|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module