JEE Main 2018 (Online) 15th April Morning ShiftQuestion 1
If $\beta$ is one of the angles between the normals to the ellipse, $x^{2}+3y^{2}=9$ at the points $(3\cos\theta,\sqrt3\sin\theta)$ and $(-3\sin\theta,\sqrt3\cos\theta)$; $\theta\in\left(0,\frac\pi2\right)$; then $\frac{2\cot\beta}{\sin2\theta}$ is equal to
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Solution
Concept. The two points have eccentric angles $\theta$ and $\theta+\frac\pi2$. Find both normal slopes and use the angle formula. Formulas used. (1) Normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(a\cos\phi,b\sin\phi)$: $ax\sec\phi-by\csc\phi=a^2-b^2$, slope $\frac ab\tan\phi$(2) $\tan\beta=\Big|\dfrac{m_1-m_2}{1+m_1m_2}\Big|$(3) $\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$Step 1 - Standard form. $x^2+3y^2=9\Rightarrow\frac{x^2}9+\frac{y^2}3=1$: $a=3$, $b=\sqrt3$, $\frac ab=\sqrt3$. Step 2 - Slopes by (1). $(3\cos\theta,\sqrt3\sin\theta)$ has angle $\theta$: $m_1=\sqrt3\tan\theta$. $(-3\sin\theta,\sqrt3\cos\theta)=\big(3\cos(\theta+\frac\pi2),\sqrt3\sin(\theta+\frac\pi2)\big)$: $m_2=\sqrt3\tan(\theta+\frac\pi2)=-\sqrt3\cot\theta$.Step 3 - Use (2). $m_1m_2=-3$, so $1+m_1m_2=-2$: $$\tan\beta=\frac{\sqrt3(\tan\theta+\cot\theta)}{2}=\frac{\sqrt3}{2}\cdot\frac{2}{\sin2\theta}=\frac{\sqrt3}{\sin2\theta}$$ (using (3)). Step 4. $\cot\beta=\frac{\sin2\theta}{\sqrt3}$, so $\frac{2\cot\beta}{\sin2\theta}=\frac{2}{\sqrt3}$. Answer: B