Let P_{1}: y = 4x^{2} and P_{2}: y = x^{2} + | vedclass

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

Question

(C) The points of intersection of $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ are found by setting $4x^{2} = x^{2} + 27$,which gives $3x^{2} = 27$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The points of intersection of $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ are found by setting $4x^{2} = x^{2} + 27$,which gives $3x^{2} = 27$,so $x = \pm 3$.The area $A_{1}$ bounded between $P_{1}$ and $P_{2}$ is $\int_{-3}^{3} ((x^{2} + 27) - 4x^{2}) dx = \int_{-3}^{3} (27 - 3x^{2}) dx = 2 \int_{0}^{3} (27 - 3x^{2}) dx = 2 [27x - x^{3}]_{0}^{3} = 2(81 - 27) = 108$ sq. units.According to the problem,the area $A_{2}$ bounded between $P_{1}$ and the line $y = \alpha x$ is $A_{2} = \frac{A_{1}}{6} = \frac{108}{6} = 18$ sq. units.The line $y = \alpha x$ intersects $P_{1}: y = 4x^{2}$ at $4x^{2} = \alpha x$,so $x(4x - \alpha) = 0$,giving $x = 0$ and $x = \frac{\alpha}{4}$.The area $A_{2} = \int_{0}^{\alpha/4} (\alpha x - 4x^{2}) dx = [\frac{\alpha x^{2}}{2} - \frac{4x^{3}}{3}]_{0}^{\alpha/4} = \frac{\alpha}{2}(\frac{\alpha^{2}}{16}) - \frac{4}{3}(\frac{\alpha^{3}}{64}) = \frac{\alpha^{3}}{32} - \frac{\alpha^{3}}{48} = \frac{3\alpha^{3} - 2\alpha^{3}}{96} = \frac{\alpha^{3}}{96}$.Setting $A_{2} = 18$,we get $\frac{\alpha^{3}}{96} = 18$,so $\alpha^{3} = 18 	imes 96 = 1728$.Thus,$\alpha = \sqrt[3]{1728} = 12$.

Let $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_{1}$,then $\alpha$ is equal to:

Similar Questions

The area enclosed by $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ is

$A$ line passing through the point $A(-2, 0)$ touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$,the parabola $P$,and the $x$-axis is:

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$,$x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$,$y^2=x$,$x=2$,and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

The area of the figure bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is ....... $sq. \text{ unit}$.

The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module