The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of ______.

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The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of pi/2. Explanation: The given curves are x3 – 3xy2 + 2 = 0 .....(i) And 3x2y – y3 – 2 = 0 ......(ii) Differentiating equation (i) w.r.t. x, we get `3x^2 - 3 * (x * 2y "dy"/"dx" + y^2 * 1) = 0 $\Rightarrow$ x^2 - 2xy "dy"?'dx" - y^2 = 0 $\Rightarrow$ 2xy "dy"/"dx" = x2 – y2 $\therefore$ "dy"/"dx" = (x^2 - y^2)/(2xy)` So slope of the curve m1 = (x^2 - y^2)/(2xy) Differentiating equation (ii) w.r.t. x, we get `3[x^2 "dy"/"dx" + y * 2x] - 3y^2 * "dy"/"dx" = 0 x^2 "dy"/"dx" + 2xy - y^2 "dy"/"dx" = 0 $\Rightarrow$ (x^3 - y^2) "dy"/"dx" = – 2xy $\therefore$ "dy"/"dx" = (-2xy)/(x^2 - y^2)` So the slope of the curve m2 = (-2xy)/(x^2 - y^2) Now m1 $\times$ m2 = `(x^2 - y^2)/(2xy) xx (-2xy)/(x^2 - y^2)` = – 1 So the angle between the curves is pi/2.

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