The two curves x³-3 x y²+2=0 and 3 x² y-y³-2=0 intersect at an angle of

Correct Answer is : π/2π / 2π/2

Correct Answer is : π/2π / 2π/2

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AP and TS EAMCET Engineering Model Paper 18

Question : 16 of 160

Marks: +1, -0

The two curves

x^3-3 x y^2+2=0

3 x^2 y-y^3-2=0

intersect at an angle of

Solution:

Given equation of curves are

x^3-3 x y^2+2=0

and

3 x^2 y-y^3-2=0

Differentiating equation (i) w.r.t.

x

, we get

\begin{array}{ll} & 3 x^2-3 x \cdot 2 y \frac{d y}{d x}-3 y^2 \cdot 1+0=0 \\\Rightarrow & \left(\frac{d y}{d x}\right)_1=\frac{x^2-y^2}{2 x y}\end{array}

Differentiating equation (ii) w.r.t.

x

, we get

\begin{array}{ll} & 3 x^2 \frac{d y}{d x}+6 \dot{x} y-3 y^2 \frac{d y}{d x}+0=0 \\\Rightarrow & \left(\frac{\dot{d y}}{d x}\right)_2=\frac{2 x y}{y^2-x^2}\end{array}

Now

\left(\frac{d y}{d x}\right)_1\left(\frac{d y}{d x}\right)_2=-1

Hence, both the curves are intersecting at right angle.

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