Question

Let the product of the focal distances of the point P(4, $ 2\sqrt{3} $) on the hyperbola H: $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $ p^2 + q^2 $ is equal to .....

Hint:

Use the properties of hyperbola, including the relationship between focal distances and the equation of the hyperbola, to solve for the required values.

Answer 1

Correct Answer: 120

Given hyperbola \( H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) \hfill (1) Point \( P(4, 2\sqrt{3}) \) lies on \( H \). Given \( PS_1 \cdot PS_2 = 32 \) Also, \( |PS_1 - PS_2| = 2a \) Since \( P(4, 2\sqrt{3}) \) lies on \( H \): \[ \frac{16}{a^2} - \frac{12}{b^2} = 1 \] \[ 16b^2 - 12a^2 = a^2b^2 \tag{2} \] \[ |PS_1 - PS_2|^2 = 4a^2 \] \[ PS_1^2 + PS_2^2 - 2 \cdot PS_1 \cdot PS_2 = 4a^2 \] \[ (ae - 4)^2 + 12 + (ae + 4)^2 + 12 - 2(32) = 4a^2 \] \[ 2a^2e^2 - 8 = 4a^2 \] \[ a^2e^2 - 4 = 2a^2 \] \[ b^2 = a^2(e^2 - 1) = 2a^2 \Rightarrow b^2 - a^2 = 4 \tag{3} \] From (2) and (3): \[ 16(a^2 + 4) - 12a^2 = a^2(a^2 + 4) \] \[ 16a^2 + 64 - 12a^2 = a^4 + 4a^2 \] \[ a^4 = 64 \Rightarrow a^2 = 8 \] \[ b^2 = 12 \] \[ p = 2b \quad \text{(length of conjugate axis)} \] \[ q = \frac{2b^2}{a} \quad \text{(length of latus rectum)} \] \[ p^2 + q^2 = 4b^2 + \frac{4b^4}{a^2} \] \[ p^2 + q^2 = 4(12) + \frac{4(12^2)}{8} \] \[ p^2 + q^2 = 48 + 72 = 120 \]