Applications Of Derivatives Question 118
Question: For the curve $ by^{2}={{(x+a)}^{3}} $ the square of subtangent is proportional to
[RPET 1999]
Options:
A) $ {{\text{(Subnormal)}}^{1/2}} $
B) Subnormal
C) $ {{\text{(Subnormal)}}^{\text{3/2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ by^{2}={{(x+a)}^{3}}\Rightarrow 2by.\frac{dy}{dx}=3{{(x+a)}^{2}}\Rightarrow \frac{dy}{dx}=\frac{3}{2by}{{(x+a)}^{2}} $ \ Subnormal = $ y\frac{dy}{dx}=\frac{3}{2b}{{(x+a)}^{2}} $ \ Subtangent = $ \frac{y}{( \frac{dy}{dx} )}=\frac{y}{\frac{3{{(x+a)}^{2}}}{2by}}=\frac{2by^{2}}{3{{(x+a)}^{2}}} $ = $ \frac{2b{{\frac{(x+a)}{b}}^{3}}}{3{{(x+a)}^{2}}}=\frac{2}{3}(x+a) $ \ (Subtangent)2 = $ \frac{4}{9}{{(x+a)}^{2}} $ and $ \frac{{{(Subtangent)}^{2}}}{Subnormal}=\frac{\frac{4}{9}{{(x+a)}^{2}}}{\frac{3}{2b}{{(x+a)}^{2}}}=\frac{8b}{27} $
therefore (Subtangent)2 = constant ´ (Subnormal). \ (Subtangent)2 µ (Subnormal).
BETA
