Question

Let $P(3\sec\theta, 2\tan\theta)$ and $Q(3\sec\phi, 2\tan\phi)$ be two points on $\frac{x^2}{9} - \frac{y^2}{4} = 1$ such that $\theta + \phi = \frac{\pi}{2}$. Then the ordinate of the intersection of the normals at $P$ and $Q$ is

• A. 13/2
• B. -13/2
• C. 5/2
• D. -5/2

Answer 1

The Correct Option is B

The standard hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with \( a = 3 \), \( b = 2 \).
So a point on it in parametric form is: \( (x, y) = (3\sec\theta, 2\tan\theta) \).

The equation of the normal to the hyperbola at \( (a\sec\theta, b\tan\theta) \) is:

\[ ax\sec\theta + by\tan\theta = a^2 + b^2 \]

Plugging in \( a = 3 \), \( b = 2 \):

Normal at \( P \): \[ 3x\sec\theta + 2y\tan\theta = 13 \tag{1} \] Normal at \( Q \): \[ 3x\sec\phi + 2y\tan\phi = 13 \tag{2} \]

We are told: \( \theta + \phi = \frac{\pi}{2} \).
Use identities:

• \( \sec\phi = \csc\theta \)
• \( \tan\phi = \cot\theta \)

Substitute into (2): \[ 3x\csc\theta + 2y\cot\theta = 13 \tag{2'} \]

Now subtract equations (1) and (2') after multiplying:

• Multiply (1) by \( \csc\theta \)
• Multiply (2') by \( \sec\theta \)

\[ \text{From (1): } 3x\sec\theta\csc\theta + 2y\tan\theta\csc\theta = 13\csc\theta \] \[ \text{From (2'): } 3x\csc\theta\sec\theta + 2y\cot\theta\sec\theta = 13\sec\theta \]

Now subtract: \[ 2y\left(\tan\theta\csc\theta - \cot\theta\sec\theta\right) = 13\left(\csc\theta - \sec\theta\right) \]

Recall: \[ \tan\theta\csc\theta = \frac{1}{\cos\theta}, \quad \cot\theta\sec\theta = \frac{1}{\sin\theta} \]

So: \[ 2y\left(\frac{1}{\cos\theta} - \frac{1}{\sin\theta}\right) = 13\left(\frac{1}{\sin\theta} - \frac{1}{\cos\theta}\right) \]

Bring all terms to one side: \[ 2y = -13 \Rightarrow y = -\frac{13}{2} \]

Answer:

\[ \boxed{-\frac{13}{2}} \]