JEE Mains · Maths · STD 12 - 6. Application of derivatives
The length of the perpendicular from the origin, on the normal to the curve, \(x^{2}+2 x y-3 y^{2}=0\) at the point \((2,2)\) is
- A \(4 \sqrt{2}\)
- B \(2 \sqrt{2}\)
- C \(2\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(x^{2}+2 x y-3 y^{2}=0\) \(\mathrm{m}_{\mathrm{N}}=\) slope of normal drawn to curve at \((2,2)\) is \(-1\) \(\mathrm{L}: \mathrm{x}+\mathrm{y}=4\) perpendicular distance of \(L\) from \((0,0)\) \(=\frac{|0+0-4|}{\sqrt{2}}=2 \sqrt{2}\)
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