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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

माना अतिपरवलय \(3 \mathrm{x}^2-4 \mathrm{y}^2=36\) पर बिन्दु \(\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)\), रेखा \(3 \mathrm{x}+2 \mathrm{y}=1\) के निकटतम है। तो \(\sqrt{2}\left(\mathrm{y}_0-\mathrm{x}_0\right)\) बराबर है:

  1. A \(-3\)
  2. B \(9\)
  3. C \(-9\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-9\)

Step-by-step Solution

Detailed explanation

\(3 x^2-4 y^2=36 \quad 3 x+2 y=1\) \(m =-\frac{3}{2}\) \(m =+\frac{\sec \theta 3}{\sqrt{12} \cdot \tan \theta}\) \(\Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2}\) \(\sin \theta=-\frac{1}{\sqrt{3}}\) \((\sqrt{12} \cdot \sec \theta, 3 \tan \theta)\)…
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