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

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(B) Given curves are $x^{3}-3xy^{2}+2=0$ $(1)$ and $3x^{2}y-y^{3}=2$ $(2)$. Differentiating $(1)$ with respect to $x$: $3x^{2}-3(y^{2}+2xyy')=0 \Right

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cut at right angle

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(B) Given curves are $x^{3}-3xy^{2}+2=0$ $(1)$ and $3x^{2}y-y^{3}=2$ $(2)$. Differentiating $(1)$ with respect to $x$: $3x^{2}-3(y^{2}+2xyy')=0 \Rightarrow x^{2}-y^{2}=2xyy' \Rightarrow y' = \frac{x^{2}-y^{2}}{2xy} = m_{1}$. Differentiating $(2)$ with respect to $x$: $3(2xy+x^{2}y')-3y^{2}y'=0 \Rightarrow 2xy+x^{2}y'-y^{2}y'=0 \Rightarrow y'(x^{2}-y^{2}) = -2xy \Rightarrow y' = -\frac{2xy}{x^{2}-y^{2}} = m_{2}$. Now,$m_{1} \cdot m_{2} = \left(\frac{x^{2}-y^{2}}{2xy}\right) \cdot \left(-\frac{2xy}{x^{2}-y^{2}}\right) = -1$. Since the product of the slopes is $-1$,the curves cut at a right angle.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{y(2x - y)}{x(x - 2y)}$

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

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