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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 52.

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 52.

Answers (1)

Answer: The value of a is 2.

Hints:

Given: The parabola y^2=16ax and  x^2=16ay cgfy

Solution:

Area of the bounded region = \frac{1024}{3}

\begin{aligned} &x=\frac{y^{2}}{16 a} \\ &y=0 ; 16 a \end{aligned}

\begin{aligned} &\text { Area }=\int_{0}^{16 a} \sqrt{16 a x} d x-\int_{0}^{16 a} \frac{x^{2}}{16 a} d x \\ &=\left[4 \sqrt{a} \frac{x^{\frac{4}{2}}}{\frac{3}{2}}\right]_{0}^{16 a}-\left[\frac{x^{3}}{48 a}\right]_{0}^{16 a} \\ &=\frac{2}{3} \times 4 \sqrt{a} \times(16 a)^{\frac{3}{2}}-0-\frac{1}{16 a} \times \frac{(16 a)^{3}}{3}-0 \end{aligned}

\begin{aligned} &A=\frac{1024}{3} \\ &\frac{1024}{3}=\frac{8 \sqrt{a}}{3} \times(16 a) \times 4 \sqrt{a}-\frac{256 a^{3}}{3} \\ &1024=256 a^{2} \\ &a^{2}=4 \\ &a=2 \end{aligned}

Therefore, the value of a is 2.

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