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  • The two curves x^3 - 3xy^2 + 2 = 0 and 3x^2y - y^3 - 2 = 0 intersect at an angle of A.

The two curves x^3 - 3xy^2 + 2 = 0and 3x^2y - y^3 - 2 = 0 intersect at an angle of
A. \frac{\pi}{4}
B. \frac{\pi}{3}
C. \frac{\pi}{2}
D. \frac{\pi}{6}

Answers (1)

Given the curve x^3 - 3xy^2 + 2 = 0and 3x^2y - y^3 - 2 = 0

Differentiate on both the sides with respect to x

\frac{\mathrm{d}\left(\mathrm{x}^{3}-3 \mathrm{xy}^{2}+2\right)}{\mathrm{dx}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}$
Apply the sum rule and also 0 is the the derivative of the constant, so it becomes
\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(3 \mathrm{xy}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(2)}{\mathrm{dx}}=0$
Apply power rule and get
\Rightarrow 3 x^{2}-3 \frac{d\left(x y^{2}\right)}{d x}+0=0$
Apply product rule and get

\begin{aligned} &\Rightarrow 3 x^{2}-3\left(x \frac{d\left(y^{2}\right)}{d x}+y^{2} \frac{d x}{d x}\right)=0\\ &\Rightarrow 3 x^{2}-3\left(x .2 y \frac{d(y)}{d x}+y^{2}\right)=0\\ &\Rightarrow 3 x^{2}-6 x y \frac{d y}{d x}-3 y^{2}=0\\ &\Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\\ &\Rightarrow \frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y}=\frac{3\left(x^{2}-y^{2}\right)}{6 x y}\\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}\\ &\text { Let it be equal to } \mathrm{m}_{\mathrm{d}_{k}} \mathrm{so}\\ &\Rightarrow \mathrm{m}_{1}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} . . \text { (i) }\\ &3 x^{2} y-y^{3}-2=0 \end{aligned}

Differentiate on both the sides with respect to x and get

\frac{d\left(3 x^{2} y-y^{3}-2\right)}{d x}=\frac{d(0)}{d x}$
Apply the sum rule and also 0 is the derivative of the constant, so
\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2} \mathrm{y}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(2)}{\mathrm{dx}}=0$

Apply power rule and get
\Rightarrow 3 \frac{d\left(x^{2} y\right)}{d x}-3 y^{2} \frac{d(y)}{d x}+0=0$

Apply product rule and get
\\\Rightarrow 3\left(x^{2} \frac{d(y)}{d x}+y \frac{d\left(x^{2}\right)}{d x}\right)-3 y^{2} \frac{d(y)}{d x}=0$ \\$\Rightarrow 3 x^{2} \frac{d y}{d x}+3 y(2 x)-3 y^{2} \frac{d y}{d x}=0$

\begin{aligned} &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)+6 \mathrm{xy}=0\\ &\Rightarrow \frac{d y}{d x}\left(3 x^{2}-3 y^{2}\right)=-6 x y\\ &\Rightarrow \frac{d y}{d x}=-\frac{6 x y}{\left(3 x^{2}-3 y^{2}\right)}\\ &\text { Let it be equal to } \mathrm{m}_{2} \mathrm{z}_{\mathrm{r}}\\ &\Rightarrow \mathrm{m}_{2}=-\frac{6 \mathrm{xy}}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)} \ldots \ldots .\\ &\text { The multiply equation (i) and (ii) and get }\\ &\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\left(\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}\right) \cdot\left(-\frac{6 \mathrm{xy}}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}\right)\\ &\Rightarrow \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\left(-\frac{3\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\left(3 \mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}\right)\\ &\Rightarrow m_{1} \cdot \mathrm{m}_{2}=-1 \end{aligned}

Since the product of the slopes is -1, it means that both the curves are intersecting at right angle i.., they are making \frac{\pi}{2} angle with each other.

So, the correct answer is option C

Posted by

infoexpert22

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