The two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2

Correct Answer is : cut at right angle

Correct Answer is : cut at right angle

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KCET 2015 Math Solved Paper

Question : 50 of 60

Marks: +1, -0

The two curves

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

3x^2y - y^3 = 2

Solution:

Given Curves,

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

→ (1)

3 x^{2} y - y^{3} = 2

→ (2)

Differentiating Eqs. (1) and (2) with respect to x, we have

3 x^{2} - 3 ( y^{2} + x ( 2 y y^{'} ) ) = 0

\Rightarrow x^{2} = y^{2} + 2 x y y^{'}

\Rightarrow y^{'} = \; \frac{x^{2} - y^{2}}{2 x y} = m_{1}

(let) → (3)

3 ( x^{2} y^{'} + 2 x y ) - 3 y^{2} y^{'} = 0

\Rightarrow x^{2} y^{'} + 2 x y - y^{2} y^{'} = 0

\Rightarrow y^{'} = - \; \frac{2 x y}{x^{2} - y^{2}} = m_{2}

(let) → (4)

From Eqs. (3) and (4), we have

m1 \cdot m_2 = -1

Hence, these curves cut at 90°

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