JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P \left( x _0, y _0\right)\) be the point on the hyperbola \(3 x ^2-4 y ^2\) \(=36\), which is nearest to the line \(3 x+2 y=1\). Then \(\sqrt{2}\left( y _0- x _0\right)\) is equal to :
- A \(-3\)
- B \(9\)
- C \(-9\)
- D \(3\)
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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