Question

Let A(2 secθ, 3 tanθ) and B(2 secΦ, 3 tanΦ) where \(\theta+\phi=\frac{\pi}{2}\) be two parts of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\). If (α,β) is the point of intersection of normals of the hyperbola at A and B, then β is equal to

• A. \(\frac{12}{3}\)
• B. \(\frac{13}{3}\)
• C. \(-\frac{12}{3}\)
• D. \(-\frac{13}{3}\)

Answer 1

The Correct Option is D

Given: The hyperbola is \[ \frac{x^2}{4} - \frac{y^2}{9} = 1 \]

Let points A and B on the hyperbola be: \[ A = (2\sec\theta, 3\tan\theta), \quad B = (2\sec\phi, 3\tan\phi) \] where \( \theta + \phi = \frac{\pi}{2} \Rightarrow \phi = \frac{\pi}{2} - \theta \)

Step 1: Derivative of the curve 

\[ \frac{d}{dx} \left( \frac{x^2}{4} - \frac{y^2}{9} \right) = 0 \Rightarrow \frac{x}{2} - \frac{2y}{9} \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{9x}{4y} \]

Step 2: Slope of normal at A

At \( A = (2\sec\theta, 3\tan\theta) \), \[ \frac{dy}{dx} = \frac{9(2\sec\theta)}{4(3\tan\theta)} = \frac{3\sec\theta}{2\tan\theta} \Rightarrow \text{slope of normal} = -\frac{2\tan\theta}{3\sec\theta} \]

Equation of normal at A:

\[ y - 3\tan\theta = -\frac{2\tan\theta}{3\sec\theta} (x - 2\sec\theta) \tag{1} \]

Step 3: Slope of normal at B

Using \( \phi = \frac{\pi}{2} - \theta \), we get: \[ \tan\phi = \cot\theta, \quad \sec\phi = \csc\theta \] So, \[ \text{slope of normal at B} = -\frac{2\cot\theta}{3\csc\theta} \]

Equation of normal at B:

\[ y - 3\cot\theta = -\frac{2\cot\theta}{3\csc\theta} (x - 2\csc\theta) \tag{2} \]

Step 4: Find the point of intersection (α, β)

Substitute equations (1) and (2) to find the coordinates where the normals intersect. Solving these equations simultaneously (by algebra or substitution), we eventually reach: \[ \beta = -\frac{13}{3} \]

Final Answer: Option (D): \[ \boxed{-\frac{13}{3}} \]