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Question

Question: Area inside the parabola \(y ^ { 2 } = 4 a x\)between the lines \(x = a\)and\(x = 4 a\)z...

Area inside the parabola y2=4axy ^ { 2 } = 4 a xbetween the lines x=ax = aandx=4ax = 4 az

A

4a24 a ^ { 2 }

B

8a28 a ^ { 2 }

C

28a2328 \frac { a ^ { 2 } } { 3 }

D

35a2335 \frac { a ^ { 2 } } { 3 }

Answer

28a2328 \frac { a ^ { 2 } } { 3 }

Explanation

Solution

We have y2=4axy ^ { 2 } = 4 a xy=2axy = 2 \sqrt { a x }

We know the equations of lines x=ax = a and x=4ax = 4 a

∴ The area inside the parabola between the lines

A=a4aydx=a4a2axdx=2aa4ax12dx=2a[x3232]a4aA = \int _ { a } ^ { 4 a } y d x = \int _ { a } ^ { 4 a } 2 \sqrt { a x } d x = 2 \sqrt { a } \int _ { a } ^ { 4 a } x ^ { \frac { 1 } { 2 } } d x = 2 \sqrt { a } \left[ \frac { x ^ { \frac { 3 } { 2 } } } { \frac { 3 } { 2 } } \right] _ { a } ^ { 4 a }

=43a12[(4a)32(a)32]=43a12a32[81]= \frac { 4 } { 3 } a ^ { \frac { 1 } { 2 } } \left[ ( 4 a ) ^ { \frac { 3 } { 2 } } - ( a ) ^ { \frac { 3 } { 2 } } \right] = \frac { 4 } { 3 } a ^ { \frac { 1 } { 2 } } a ^ { \frac { 3 } { 2 } } [ 8 - 1 ] =283a2= \frac { 28 } { 3 } a ^ { 2 }.