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

The hyperbola equation is given as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). The sum of the focal distances for point \( P(4, 3) \) is \( 8\sqrt{\frac{5}{3}} \).
Step 1: Determine the eccentricity from the given conditions:
\[2e = 8\sqrt{\frac{5}{3}} \quad \Rightarrow \quad e = \sqrt{\frac{5}{3}}\]
Step 2: Apply the hyperbola's focal distance relationship:
\[b^2 = a^2 \left( \left( \sqrt{\frac{5}{3}} \right)^2 - 1 \right)\] \[b^2 = a^2 \left( \frac{5}{3} - 1 \right) = a^2 \times \frac{2}{3}\]
Step 3: Establish 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 the 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: Calculate the length of the latus rectum \( \ell \):
\[\ell = \frac{2b^2}{a} = \frac{2 \times \frac{2a^2}{3}}{a} = \frac{4a^2}{3a} = \frac{4a}{3}\] With \( a^2 = \frac{5}{2} \), we have \( a = \sqrt{\frac{5}{2}} \). Therefore:\[\ell = \frac{4\sqrt{\frac{5}{2}}}{3}\]
Step 5: Compute \( m \), the product of the focal distances:
\[m = (e \cdot a) \left( e - a \right)\] Substituting \( e = \sqrt{\frac{5}{3}} \) 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, compute \( 9\ell^2 + 6m \):
\[9\ell^2 + 6m = 36 \times \frac{5}{9} + 6 \times 145 \quad \Rightarrow \quad 9\ell^2 + 6m = 185\]
The resulting value is \( 185 \).