Question

Let the sum of the focal distances of the point $ P(4, 3) $ on the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be $ 8\sqrt{\frac{5}{3}} $. If for $ H $, the length of the latus rectum is $ \ell $ and the product of the focal distances of the point $ P $ is $ m $, then $ 9\ell^2 + 6m $ is equal to:

• A. 184
• B. 186
• C. 185
• D. 187

Hint:

In hyperbola problems, use the relationships between \( a^2 \), \( b^2 \), and \( e^2 \) to solve for the unknowns. The focal distance and latus rectum can be computed from these relations.

Answer 1

The Correct Option is C

We are given the hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), and the sum of the focal distances of the point \( P(4, 3) \) is \( 8\sqrt{\frac{5}{3}} \).
Step 1: Find the distance using the given conditions:
\[ 2e = 8\sqrt{\frac{5}{3}} \quad \Rightarrow \quad e = \sqrt{\frac{5}{3}} \]
Step 2: Use the relationship for the focal distance in a hyperbola:
\[ b^2 = a^2 \left( \left( \sqrt{\frac{5}{3}} \right)^2 - 1 \right) \] \[ b^2 = a^2 \left( \frac{5}{3} - 1 \right) = a^2 \times \frac{2}{3} \]
Step 3: Use the relationship between \( a^2 \) and \( b^2 \):
\[ \frac{16}{a^2} - \frac{9}{b^2} = 1 \] Substitute \( b^2 = \frac{2a^2}{3} \) into this equation: \[ \frac{16}{a^2} - \frac{9}{\frac{2a^2}{3}} = 1 \quad \Rightarrow \quad \frac{16}{a^2} - \frac{27}{2a^2} = 1 \] \[ \frac{32}{2a^2} - \frac{27}{2a^2} = 1 \quad \Rightarrow \quad \frac{5}{2a^2} = 1 \quad \Rightarrow \quad a^2 = \frac{5}{2} \]
Step 4: Now, calculate the length of the latus rectum \( \ell \):
\[ \ell = \frac{2b^2}{a} = \frac{2 \times \frac{2a^2}{3}}{a} = \frac{4a^2}{3a} = \frac{4a}{3} \] Substitute \( a^2 = \frac{5}{2} \), we get \( a = \sqrt{\frac{5}{2}} \), so: \[ \ell = \frac{4\sqrt{\frac{5}{2}}}{3} \]
Step 5: Calculate \( m \) (the product of the focal distances):
\[ m = (e \cdot a) \left( e - a \right) \] Substitute \( e = \sqrt{\frac{5}{3}} \) and \( a = \sqrt{\frac{5}{2}} \): \[ m = \sqrt{\frac{5}{3}} \times \sqrt{\frac{5}{2}} \times \left( \sqrt{\frac{5}{3}} - \sqrt{\frac{5}{2}} \right) \]
Step 6: Finally, calculate \( 9\ell^2 + 6m \):
\[ 9\ell^2 + 6m = 36 \times \frac{5}{9} + 6 \times 145 \quad \Rightarrow \quad 9\ell^2 + 6m = 185 \]
Thus, the correct answer is \( 185 \).

Answer 2

Step 1: Solve for \( e \): 

\[ ex + a + ex - a = 8\sqrt{\frac{5}{3}}. \] Simplifying, we get: \[ 2ex = 8\sqrt{\frac{5}{3}}. \] Thus: \[ e \times 4 = 8\sqrt{\frac{5}{3}}, \] which gives: \[ e = \sqrt{\frac{5}{3}}. \]

Step 2: Solve for \( b^2 \):

\[ b^2 = a^2 \left( \left( \frac{\sqrt{5}}{3} \right)^2 - 1 \right). \] Simplifying this: \[ b^2 = a^2 \left( \frac{5}{3} - 1 \right) = \frac{5}{3}a^2. \] Thus: \[ a^2 = \frac{3}{2}, \quad b^2 = \frac{5}{3}. \]

Step 3: Solve for \( \ell \):

\[ \ell = 2b^2. \] Substituting \( b^2 = \frac{5}{3} \): \[ \ell = 2 \times \frac{5}{3} = \frac{10}{3}. \]

Step 4: Solve for \( g\ell^2 \):

\[ g\ell^2 = 36 \times \frac{25}{9} \times 2 = 40. \]

Step 5: Solve for \( m \):

\[ m = (ex + a)(ex - a). \] Substituting the values of \( e \) and solving: \[ m = e^2a^2 - a^2 = \frac{5}{3} \times 16 - \frac{5}{3} = \frac{145}{6}, \] which gives: \[ m = 6m + 145. \]

Step 6: Final Calculation:

\[ 40 + 145 = 185. \]