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KCET2015MathematicsDeterminants

The two curves ( x³-3 x y²+2=0 ) and ( 3 x² y-y³=2 )

Options

  1. Atouch each other
  2. Bcut at right angle
  3. Ccut at angle ( 3 )
  4. Dcut at angle ( 4 )

Correct answer

B. cut at right angle

Step-by-step solution

Given Curves, x³-3 x y²+2=0 (1)3 x² y-y³=2 (2) Differentiating Eqs. (1) and (2) with respect to x, we have 3 x²-3 (y²+x(2 y y) )=0 x²=y²+2 x y y y= x²-y² 2 x y =m₁ (let) (3)3 (x² y+2 x y )-3 y² y=0 x² y+2 x y-y² y=0 y=- 2 x y x²-y² =m₂ (let) (4) From Eqs. (3) and (4), we have m 1 m₂=-1 Hence, these curves cut at 90^

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