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  • Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 3

Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 3

Answers (1)

\frac{1}{24a}

Hint:

Find the latus rectum and its intersection points with the parabola and then integrate the expression to find the area enclosed by the curves.

Given:

               x^{2}=4ay

Solution:

First, we find the latus rectum and its intersection points with the parabola

                   x^{2}=4ay     … (i)

                     y=\frac{x^{2}}{4a}   … (ii)

Comparing it with the standard from of a parabola

               Y^{2}=4aX^{2}

Where (0,A) is the coordinate of the focus of the parabola and the latus rectum passes through this point and is perpendicular to the axis of symmetry

Therefore, the equation of the latus rectum is y=a

Comparing equation (i) and (ii)

               \frac{1}{4a}=4A

A=\frac{1}{16a}

Intersection point

               \begin{aligned} &\frac{1}{16 a}=\frac{x^{2}}{4 a} \\ &x^{2}=\frac{1}{4} \\ &x=\pm \frac{1}{2} \end{aligned}

Now, Integrating the expression to find the area enclosed by the curves

Since, the latus rectum is above the parabola in the Cartesian plane, the expression will be

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

\begin{aligned} &=\frac{1}{16 a}-\frac{1}{48 a} \\ &=\frac{1}{24 a} \end{aligned}

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