Definite Integration Question 308
Question: The area enclosed by the parabola $ y^{2}=4ax $ and the straight line $ y=2ax, $ is
[MP PET 1993]
Options:
A) $ \frac{a^{2}}{3} $ sq. unit
B) $ \frac{1}{3a^{2}} $ sq. unit
C) $ \frac{1}{3a} $ sq. unit
D) $ \frac{2}{3a} $ sq. unit
Show Answer
Answer:
Correct Answer: C
Solution:
The points of intersection of the parabola $ y^{2}=4ax $ and the chord $ y=2ax $ is obtained by solving these equations simultaneously. $ y^{2}=4ax,y=2ax\Rightarrow {{(2ax)}^{2}}=4ax $
therefore $ x[4a^{2}x-4a]=0 $
$ \Rightarrow 4ax[ax-1]=0 $
therefore $ x=0 $ or $ x=\frac{1}{a} $ Also $ x=0\Rightarrow y=0 $ and $ x=\frac{1}{a} $
therefore $ y=\pm 2 $
Hence the required points are (0,0) and $ [ \frac{1}{a},2 ] $ . Now required area $ =\int_0^{1/a}{[\sqrt{4ax}-2ax}]dx=\frac{1}{3a}sq $ . unit.
BETA
