Let the product of the focal distances of the point $\mathbf{P}(4,2 \sqrt{3})$ on the hyperbola $\mathrm{H}: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ 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__________

<p>To find $ p^2 + q^2 $ for the hyperbola $\mathrm{H}: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, given that the product of the focal distances from a point $ \mathbf{P}(4, 2\sqrt{3}) $ to the foci is 32, we follow these steps:</p> <p><p><strong>Verify the Point on the Hyperbola:</strong></p> <p>The point $ P(4, 2\sqrt{3}) $ satisfies the hyperbola equation:</p> <p>$ \frac{16}{a^2} - \frac{12}{b^2} = 1 \quad \ldots \text{(i)} $</p></p> <p><p><strong>Product of Focal Distances:</strong></p> <p>The distances from $ P $ to the foci $ S $ and $ S' $ are:</p> <p>$ SP = e\left(4 - \frac{a}{e}\right), \quad S'P = e\left(4 + \frac{a}{e}\right) $</p> <p>Therefore, the product:</p> <p>$ SP \cdot S'P = 16e^2 - a^2 = 32 $</p></p> <p><p><strong>Relating $ e $, $ a $, and $ b $:</strong></p> <p>Using the identity for eccentricity:</p> <p>$ e^2 = 1 + \frac{b^2}{a^2} $</p> <p>Substituting in the product equation gives:</p> <p>$ 16\left(1 + \frac{b^2}{a^2}\right) - a^2 = 32 \quad \ldots \text{(ii)} $</p></p> <p><p><strong>Solving for $ a^2 $ and $ b^2 $:</strong></p> <p>From equations (i) and (ii), solve for $ a^2 $ and $ b^2 $:</p> <p>$ a^2 = 8, \quad b^2 = 12 $</p></p> <p><p><strong>Calculate $ p^2 + q^2 $:</strong></p> <p>The conjugate axis length is $ p = 2b $ and the latus rectum is $ q = \frac{2b^2}{a} $. Thus:</p> <p>$ p^2 + q^2 = (2b)^2 + \left(\frac{2b^2}{a}\right)^2 $</p> <p>Substituting $ b^2 = 12 $ and $ a^2 = 8 $ gives:</p> <p>$ p^2 + q^2 = 4b^2 + \frac{4b^4}{a^2} = 48 + \frac{4 \times 144}{8} $</p> <p>$ p^2 + q^2 = 48 \cdot \left(1 + \frac{12}{8}\right) = 120 $</p></p> <p>Therefore, $ p^2 + q^2 = 120 $.</p>