The two curves x^{3}-3xy^{2}+2=0 and 3x^{2}y- | vedclass

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(B) Given curves are: $x^{3}-3xy^{2}+2=0 \quad (1)$ $3x^{2}y-y^{3}=2 \quad (2)$ Differentiating equation $(1)$ with respect to $x$: $3x^{2} - 3(y^{2}

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Cut each other at right angle

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(B) Given curves are: $x^{3}-3xy^{2}+2=0 \quad (1)$ $3x^{2}y-y^{3}=2 \quad (2)$ Differentiating equation $(1)$ with respect to $x$: $3x^{2} - 3(y^{2} + x(2yy')) = 0$ $x^{2} - y^{2} - 2xyy' = 0$ $y' = \frac{x^{2}-y^{2}}{2xy} = m_{1}$ Differentiating equation $(2)$ with respect to $x$: $3(2xy + x^{2}y') - 3y^{2}y' = 0$ $2xy + x^{2}y' - y^{2}y' = 0$ $y' = \frac{-2xy}{x^{2}-y^{2}} = m_{2}$ Now,calculating the product of slopes: $m_{1} 	imes m_{2} = \left(\frac{x^{2}-y^{2}}{2xy}ight) 	imes \left(\frac{-2xy}{x^{2}-y^{2}}ight) = -1$ Since the product of the slopes is $-1$,the two curves intersect at a right angle.

The two curves $x^{3}-3xy^{2}+2=0$ and $3x^{2}y-y^{3}=2$:

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