Definite Integration Question 232
Question: If area bounded by the curves $ y^{2}=4ax $ and $ y=mx $ is $ a^{2}/3, $ , then the value of $ m $ is
Options:
A) 2
B) $ -2 $
C) $ \frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The two curves $ y^{2}=4ax $ and $ y=mx $ intersect at $ ( \frac{4a}{m^{2}},\frac{4a}{m} ) $ and the area enclosed by the two curves is given by $ \int_0^{4a/m^{2}}{(\sqrt{4ax}-mx)}dx $ .
$ \therefore \int_0^{4a/m^{2}}{(\sqrt{4ax}-mx)}dx=\frac{a^{2}}{3} $
therefore $ \frac{8}{3}\frac{a^{2}}{m^{3}}=\frac{a^{2}}{3}\Rightarrow m^{3}=8\Rightarrow m=2 $ .
BETA
