Definite Integration Question 128
Question: If the area enclosed by $y^2=4 a x$ and the line $y=a x$ is $\frac{1}{3}$ sq. unit, then the roots of the equation $x^2+2 x=a$, are
Options:
A) -4 and 2
B) 2 and 4
C) -2 and -4
D) 8 and -8
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y=\int_0^{4/a}{(a\cdot x-\sqrt{4a\cdot x})dx} $
$ \frac{1}{3}=\int_0^{4/a}{axdx-\int_0^{4/a}{\sqrt{4ax}dx}} $
$ \frac{1}{3}=[ \frac{ax^{2}}{2} ]_0^{4/a}-2[ \frac{{{(4ax)}^{3/2}}}{3} ]_0^{4/a} $
$ \frac{1}{3}=\frac{\frac{16a}{a^{2}}}{2}-\frac{2}{3}[ 4a{{( \frac{4}{a} )}^{3/2}} ],a=8 $ .
Putting the value of a in $ x^{2}+2x-a=0 $ , we get its roots i.e., -4 and 2.
BETA
