Question
The two curves $x^{3}-3 x y^{2}+2=0$ and $3 x^{2} y-y^{3}-2=0$(a) cut at right angles(b) touch each other(c) cut at an angle $\pi 3$(d) cut at an angle $\pi / 4$
Step 1
The first equation is $x^{3}-3 x y^{2}+2=0$. Differentiating it with respect to $x$, we get $3x^{2}-3y^{2}-6xy\frac{dy}{dx}=0$. From this, we can solve for $\frac{dy}{dx}$ to get the slope of the tangent at any point on the curve. We get Show more…
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