AP EAMCET · Maths · Application of Derivatives
If \(\beta\) is an angle between the normals drawn to the curve \(x^2+3 y^2=9\) at the points \((3 \cos \theta, \sqrt{3} \sin \theta)\) and \((-3 \sin \theta, \sqrt{3} \cos \theta), \theta \in\left(0, \frac{\pi}{2}\right)\), then
- A \(\tan \beta=\frac{1}{\sqrt{3}} \sec 2 \theta\)
- B \(\cot \beta=\sqrt{3} \operatorname{cosec} 2 \theta\)
- C \(\sqrt{3} \cot \beta=\sin 2 \theta\)
- D \(\cot \beta=\frac{1}{\sqrt{2}} \sec 2 \theta\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3} \cot \beta=\sin 2 \theta\)
Step-by-step Solution
Detailed explanation
Given curve: \(x^2+3 y^2=9\) Differentiate implicitly: \(2x+6y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{x}{3y}\) Slope of normal \(m_N = -\frac{1}{\frac{dy}{dx}} = \frac{3y}{x}\) Slope of normal at \(P_1(3 \cos \theta, \sqrt{3} \sin \theta)\):…
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