Applications Of Derivatives Question 61
Question: The two curves $ x^{3}-3xy^{2}+2=0and3x^{2}y-y^{3}=2 $
Options:
A) Cuts at right angle
B) Touch each other
C) Cut at an angle $ \frac{\pi }{3} $
D) Cut at an angle $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Two curves cuts at right angle if product of their slopes is -1. Two gives curves are $ x^{3}-3xy^{2}+2=0 $ - (i) and $ 3x^{2}y-y^{3}-2=0 $ - (ii) Differentiate equ. (i), $ 3x^{2}-3[ y^{2}+2xy\frac{dy}{dx} ]=0 $
$ \Rightarrow 3(x^{2}-y^{2})=6xy\frac{dy}{dx} $
$ \Rightarrow m_1=\frac{dy}{dx}=\frac{3(x^{2}-y^{2})}{6xy} $ Differentiate eq. (ii), $ 3x^{2}y-y^{3}-2=0 $
$ \Rightarrow 3[ x^{2}\frac{dy}{dx}+2xy ]-3y^{2}\frac{dy}{dx}=0 $
$ \Rightarrow x^{2}\frac{dy}{dx}+2xy-y^{2}\frac{dy}{dx}=0 $
$ \Rightarrow (x^{2}-y^{2})\frac{dy}{dx}=-2xy $
$ \Rightarrow m_2=\frac{dy}{dx}=\frac{-2xy}{(x^{2}-y^{2})} $
$ \therefore m_1\times m_2=\frac{(x^{2}-y^{2})}{2xy}\times \frac{-2xy}{(x^{2}-y^{2})} $
$ \Rightarrow m_1\times m_2=-1 $ i.e., curves cuts at right angle.
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