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
- A fourth
- B never again
- C second
- D third
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…
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