Definite Integration Question 186
Question: If the area enclosed by $ y^{2}=4ax $ and line $ y=ax $ is 1/3 sq. units , then the area enclosed by $ y=4x $ with same parabola is
Options:
A) 8 sq. units
B) 4 sq. units
C) 4/3 sq. units
D) 8/3 sq. units
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Point of intersection of $ y^{2}=4ax $ and $ y=ax $ are (0, 0) and $ ( \frac{4}{a},4 ) $ Given $ \int\limits_0^{4}{[ \frac{y}{a}-\frac{y^{2}}{4a} ]dy=\frac{1}{3}} $
$ \Rightarrow \frac{8}{a}-\frac{1}{12a}\times 64=\frac{1}{3}\Rightarrow \frac{8}{3a}=\frac{1}{3}\Rightarrow a=8 $ So, the parabola is $ y^{2}=32x $ Area enclosed by $ y=4x $ is $ \int\limits_0^{8}{[ \frac{y}{4}-\frac{y^{2}}{32} ]dy=[ \frac{y^{2}}{8}-\frac{y^{3}}{96} ]}_0^{8}=\frac{8}{3} $
BETA
