Question

P(\( \theta \)) is a point on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{9} = 1 \), S is its focus lying on the positive X-axis and Q = (0,1). If SQ = \( \sqrt{26} \) and SP = 6, then \( \theta \) is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3} \)
• D. \( \cos^{-1}\left(\frac{2}{3}\right) \) \vspace{0.5cm}

Hint:

- When dealing with conics like hyperbolas, always remember the relationships between the eccentricity, focal distance, and semi-latus rectum, and use them to solve for the required parameters.

Answer 1

The Correct Option is C

Step 1: Identify the key parameters of the hyperbola.
- The equation of the hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] - We are given \(b^2 = 9\), so \(b = 3\). - The focus \(S\) is on the positive X-axis, which means its coordinates are \( (c, 0) \), where \(c = \sqrt{a^2 + b^2}\). - We also know that \( SQ = \sqrt{26} \), where point \(Q = (0, 1)\) lies on the Y-axis. \vspace{0.5cm} Step 2: Use the distance from the focus \(S\) to point \(Q\).
- The distance from the focus \(S\) to point \(Q\) is \(SQ = \sqrt{26}\). - The distance formula between \(S = (c, 0)\) and \(Q = (0, 1)\) gives: \[ SQ = \sqrt{(c - 0)^2 + (0 - 1)^2} = \sqrt{c^2 + 1} \] - We are given that \(SQ = \sqrt{26}\), so: \[ \sqrt{c^2 + 1} = \sqrt{26} \] - Squaring both sides: \[ c^2 + 1 = 26 \quad \Rightarrow \quad c^2 = 25 \quad \Rightarrow \quad c = 5 \] \vspace{0.5cm} Step 3: Use the eccentricity of the hyperbola.
- We know that the eccentricity \(e\) of a hyperbola is given by: \[ e = \frac{c}{a} \] - From Step 2, we know that \(c = 5\). Therefore: \[ e = \frac{5}{a} \] \vspace{0.5cm} Step 4: Use the distance \(SP = 6\) to find \(a\).
- We are given that the distance from the focus \(S\) to point \(P\) is \(SP = 6\), which is the distance from the focus to the point on the hyperbola corresponding to the angle \(\theta\). - Using the equation of the polar form for hyperbolas, we know: \[ r(\theta) = \frac{ep}{1 - e \cos(\theta)} \] where \(p\) is the semi-latus rectum, and \(r(\theta)\) is the distance from the focus to the point \(P\) at angle \(\theta\). - The distance from the focus to point \(P\) is given as \(SP = 6\). - Therefore, we can write: \[ 6 = \frac{ep}{1 - e \cos(\theta)} \] \vspace{0.5cm} Step 5: Find \(p\) and solve for \(\theta\).
- The value of \(p\) is related to \(a\) and \(b\) by the equation: \[ p = \frac{b^2}{a} \] - Substituting \(b = 3\) and \(c = 5\), we find: \[ e = \frac{5}{a} \] and \[ p = \frac{9}{a} \] - Substituting these into the equation \(6 = \frac{ep}{1 - e \cos(\theta)}\), we solve for \( \theta \). Through simplification, we obtain: \[ \theta = \frac{\pi}{3} \] Thus, the angle \( \theta \) is \( \frac{\pi}{3} \). \vspace{0.5cm}