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JEE Mains · Maths · STD 12 - 6. Application of derivatives

Normal of  the curve \({x^2} + 2xy - 3{y^2} = 0\) at point \((1,1)\) intersect curve again at  which quadrant

  1. A fourth 
  2. B never again
  3. C second
  4. D third
Verified Solution

Answer & Solution

Correct Answer

(A) fourth 

Step-by-step Solution

Detailed explanation

\({x^2} + 3xy - xy - 3{y^2} = 0\) \(x\left( {x + 3y} \right) - y\left( {x + 3y} \right) = 0\) \(\left( {x + 3y} \right)\left( {x - y} \right) = 0\) Equation of normal is \(\left( {y - 1} \right) = - 1\left( {x - 1} \right)\) \( \Rightarrow x + y = 2\) It intersects \(x+3y=0\) at…