JEE Mains · Maths · STD 12 - 6. Application of derivatives
If \(\beta \) is one of the angles between the normals to the ellipse, \(x^2 + 3y^2 = 9\) at the points \(\left( {3\cos \theta ,\sqrt {3\,} \sin \theta } \right)\) and \(\left( { - 3\sin \,\theta ,\sqrt 3 \,\cos \theta } \right); \in \left( {0,\frac{\pi }{2}} \right)\); then \(\frac{{2\,\cot \beta }}{{\sin \,2\theta }}\) is equal to
- A \(\sqrt 2 \)
- B \(\frac{2}{{\sqrt 3 }}\)
- C \(\frac{1}{{\sqrt 3 }}\)
- D \(\frac{{\sqrt 3 }}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{{\sqrt 3 }}\)
Step-by-step Solution
Detailed explanation
Since, \({x^2} + 3{y^2} = 9\) \( \Rightarrow 2x + 6y\frac{{dy}}{{dx}} = 0\) \( \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - x}}{{3y}}\) Slope of normal is \( - \frac{{dx}}{{dy}} = \frac{{3y}}{x}\)…
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