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CUET · MATHS · PYQ PAPER 2023

Two curves \(x^3-3 x y^2+2=0\) and \(3x^2 y-y^3=2\) :

  1. A Touch each other
  2. B Cut at right angle
  3. C Cut at an angle \(\frac{\pi}{3}\)
  4. D Cut at an angle \(\frac{\pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) Cut at right angle

Step-by-step Solution

Detailed explanation

\(m_1 = \frac{d}{dx}(x^3-3 x y^2+2=0)\) \(3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{x^2 - y^2}{2xy}\) \(m_2 = \frac{d}{dx}(3x^2 y-y^3=2)\) \(6xy + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{6xy}{3y^2 - 3x^2} = \frac{2xy}{y^2 - x^2}\)…