Question

If \( \beta \) is an angle between the normals drawn to the curve \( x^2+3y^2=9 \) 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. \( \cot\beta = \frac{1}{\sqrt{2}} \sec 2\theta \)

Hint:

Equation of ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Parametric point \( (a\cos\phi, b\sin\phi) \). Slope of tangent \( \frac{dy}{dx} = -\frac{b^2x}{a^2y} \). At \( (a\cos\phi, b\sin\phi) \), \( m_T = -\frac{b\cos\phi}{a\sin\phi} \). Slope of normal \( m_N = \frac{a\sin\phi}{b\cos\phi} = \frac{a}{b}\tan\phi \). For \(x^2+3y^2=9 \implies \frac{x^2}{9}+\frac{y^2}{3}=1\), so \(a=3, b=\sqrt{3}\). \(m_N = \frac{3}{\sqrt{3}}\tan\phi = \sqrt{3}\tan\phi \). Angle \( \beta \) between lines with slopes \(m_1, m_2\) is \( \tan\beta = |\frac{m_1-m_2}{1+m_1m_2}| \). Identity: \( \tan\theta+\cot\theta = \frac{1}{\sin\theta\cos\theta} = \frac{2}{\sin 2\theta} \).

Answer 1

The Correct Option is C

The curve is an ellipse \( \frac{x^2}{9} + \frac{y^2}{3} = 1 \).
Here \( a^2=9, b^2=3 \).
The equation of the normal to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at a point \( (x_0, y_0) \) is \( \frac{a^2x}{x_0} - \frac{b^2y}{y_0} = a^2-b^2 \).
Or, at parametric point \( (a\cos\phi, b\sin\phi) \), the normal is \( ax\sec\phi - by\operatorname{cosec}\phi = a^2-b^2 \).
Point 1: \( P_1 = (3\cos\theta, \sqrt{3}\sin\theta) \).
This is \( (a\cos\theta, b\sin\theta) \) with \( a=3, b=\sqrt{3} \).
Slope of tangent at \(P_1\): Differentiate \( x^2+3y^2=9 \implies 2x+6y\frac{dy}{dx}=0 \implies \frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y} \).
At \(P_1\), \( m_{T1} = -\frac{3\cos\theta}{3(\sqrt{3}\sin\theta)} = -\frac{\cos\theta}{\sqrt{3}\sin\theta} \).
Slope of normal at \(P_1\), \( m_{N1} = -1/m_{T1} = \frac{\sqrt{3}\sin\theta}{\cos\theta} = \sqrt{3}\tan\theta \).
Point 2: \( P_2 = (-3\sin\theta, \sqrt{3}\cos\theta) \).
Let's check if this point is \( (a\cos\phi, b\sin\phi) \).
\( a\cos\phi = -3\sin\theta \implies 3\cos\phi = -3\sin\theta \implies \cos\phi = -\sin\theta = \cos(\pi/2+\theta) \).
So \( \phi = \pi/2+\theta \).
\( b\sin\phi = \sqrt{3}\cos\theta \implies \sqrt{3}\sin\phi = \sqrt{3}\cos\theta \implies \sin\phi = \cos\theta = \sin(\pi/2-\theta) \).
Also \( \sin(\pi/2+\theta) = \cos\theta \).
This matches.
So, point \(P_2\) corresponds to parameter \( \phi = \pi/2+\theta \).
Slope of normal at \(P_2\), \( m_{N2} = \sqrt{3}\tan\phi = \sqrt{3}\tan(\pi/2+\theta) = \sqrt{3}(-\cot\theta) = -\sqrt{3}\cot\theta \).
Angle \( \beta \) between normals: \( \tan\beta = \left|\frac{m_{N1}-m_{N2}}{1+m_{N1}m_{N2}}\right| \).
\( m_{N1}m_{N2} = (\sqrt{3}\tan\theta)(-\sqrt{3}\cot\theta) = -3(\tan\theta\cot\theta) = -3(1) = -3 \).
\( m_{N1}-m_{N2} = \sqrt{3}\tan\theta - (-\sqrt{3}\cot\theta) = \sqrt{3}(\tan\theta+\cot\theta) \) \[ = \sqrt{3}\left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right) = \sqrt{3}\left(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right) = \sqrt{3}\frac{1}{\frac{1}{2}\sin 2\theta} = \frac{2\sqrt{3}}{\sin 2\theta} \] \[ \tan\beta = \left|\frac{2\sqrt{3}/\sin 2\theta}{1+(-3)}\right| = \left|\frac{2\sqrt{3}/\sin 2\theta}{-2}\right| = \left|-\frac{\sqrt{3}}{\sin 2\theta}\right| \] Since \( \theta \in (0, \pi/2) \), \( 2\theta \in (0, \pi) \), so \( \sin 2\theta>0 \).
\[ \tan\beta = \frac{\sqrt{3}}{\sin 2\theta} \] We need to match this with options.
This means \( \cot\beta = \frac{\sin 2\theta}{\sqrt{3}} \implies \sqrt{3}\cot\beta = \sin 2\theta \).
This matches option (3).