The two curves x³-3xy²+2=0 and 3x²y-y³=2

Correct Answer is : Cut each other at right angle

Correct Answer is : Cut each other at right angle

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KCET 2016 Math Solved Paper

Question : 31 of 60

Marks: +1, -0

The two curves

x^3-3xy^2+2=0

3x^2y-y^3=2

Solution:

Given curves,→

x^{3}-3xy^{2}+2=0

→ (1)

3x^{2}y-y^{3}=2

→ (2)

Differentiating Eqs. (1) and (2) both the sides with respect to x, we get

3x^{2}-3(y^{2}+x(2yy'))=0

\Rightarrow x^{2}=y^{2}+2xyy'

\Rightarrow y'=\frac{x^{2}-y^{2}}{2xy}=m_{1}

(Let) → (3)

3(x^{2}y'+2xy)3y^{2}y'=0

\Rightarrow x^{2}y'+2xy-y^{2}y'=0

\Rightarrow y'=\frac{-2xy}{x^{2}-y^{2}}=m_{2}

(Let) → (4)

From Eqs. (3) and (4), we get

m_{1}m_{2}=-1

Hence, the two curves cut at an angle

\frac{\pi}{2}

i.e. right angle

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