Question

Let X-axis be the transverse axis and Y-axis be the conjugate axis of a hyperbola H. Let \( x^2 + y^2 = 16 \) be the director circle of H. If the perpendicular distance from the centre of H to its latus rectum is \( \sqrt{34} \), then \( a + b = \):

• A. \( 8 \)
• B. \( 9 \)
• C. \( 5 \)
• D. \( 7 \)

Hint:

Remember the standard equation of a hyperbola, its director circle, foci, and latus rectum. The perpendicular distance from the centre to the latus rectum \( x = \pm ae \) is \( ae \).

Answer 1

The Correct Option is A

Step 1: Write the equation of the hyperbola and its director circle.
Since the X-axis is the transverse axis and the Y-axis is the conjugate axis, the equation of the hyperbola is of the form: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] The equation of the director circle of this hyperbola is given by \( x^2 + y^2 = a^2 - b^2 \). We are given that the director circle is \( x^2 + y^2 = 16 \). Comparing the two equations of the director circle, we have: \[ a^2 - b^2 = 16 \quad \cdots (1) \]
Step 2: Use the information about the perpendicular distance from the centre to the latus rectum. The centre of the hyperbola is \( (0, 0) \). The equation of the latus rectum is \( x = \pm \frac{b^2}{a} \). The foci are at \( (\pm ae, 0) \), where \( e \) is the eccentricity. The equations of the latus recta are \( x = \pm ae \). The length of the latus rectum is \( \frac{2b^2}{a} \). The perpendicular distance from the centre \( (0, 0) \) to the latus rectum \( x = ae \) is \( \frac{|0 - ae|}{\sqrt{1^2 + 0^2}} = |ae| = ae \) (since \( a>0, e>1 \)). So, we have \( ae = \sqrt{34} \quad \cdots (2) \). We know that \( b^2 = a^2 (e^2 - 1) \), which implies \( e^2 = 1 + \frac{b^2}{a^2} \).
Step 3: Solve the system of equations. From equation (2), \( a^2 e^2 = 34 \). Substituting the expression for \( e^2 \): \[ a^2 \left( 1 + \frac{b^2}{a^2} \right) = 34 \] \[ a^2 + b^2 = 34 \quad \cdots (3) \] Now we have a system of two equations with two variables \( a^2 \) and \( b^2 \): 1. \( a^2 - b^2 = 16 \) 2. \( a^2 + b^2 = 34 \) Adding equations (1) and (3): \[ (a^2 - b^2) + (a^2 + b^2) = 16 + 34 \] \[ 2a^2 = 50 \implies a^2 = 25 \implies a = 5 \quad \text{(since } a>0\text{)} \] \text{Subtracting equation (1) from equation (3):} \[ (a^2 + b^2) - (a^2 - b^2) = 34 - 16 \] \[ 2b^2 = 18 \implies b^2 = 9 \implies b = 3 \quad \text{(since } b>0\text{)} \]
Step 4: Calculate \( a + b \). \[ a + b = 5 + 3 = 8 \] This matches option (1).