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
Solution:
$$H: \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$
$$P(4,2\sqrt{3}) \text{ lies on } H$$
$$\frac{16}{a^2} – \frac{12}{b^2} = 1 \Rightarrow 16b^2 – 12a^2 = a^2b^2 \quad \dots (1)$$
$$\text{Let foci be } S_1(ae, 0) \text{ and } S_2(-ae, 0)$$
$$PS_1 \cdot PS_2 = 32$$
$$\text{By property of hyperbola, } |PS_1 – PS_2| = 2a$$
$$(PS_1 – PS_2)^2 = 4a^2$$
$$PS_1^2 + PS_2^2 – 2PS_1 PS_2 = 4a^2$$
$$[(4-ae)^2 + (2\sqrt{3}-0)^2] + [(4+ae)^2 + (2\sqrt{3}-0)^2] – 2(32) = 4a^2$$
$$(16 – 8ae + a^2e^2 + 12) + (16 + 8ae + a^2e^2 + 12) – 64 = 4a^2$$
$$2a^2e^2 + 56 – 64 = 4a^2$$
$$2a^2e^2 – 8 = 4a^2$$
$$a^2e^2 – 4 = 2a^2$$
$$\text{We know } a^2e^2 = a^2 + b^2$$
$$a^2 + b^2 – 4 = 2a^2$$
$$b^2 – a^2 = 4 \Rightarrow b^2 = a^2 + 4$$
$$\text{Substitute } b^2 \text{ in equation } (1):$$
$$16(a^2 + 4) – 12a^2 = a^2(a^2 + 4)$$
$$16a^2 + 64 – 12a^2 = a^4 + 4a^2$$
$$4a^2 + 64 = a^4 + 4a^2$$
$$a^4 = 64 \Rightarrow a^2 = 8$$
$$b^2 = 8 + 4 = 12$$
$$\text{Length of conjugate axis } p = 2b \Rightarrow p^2 = 4b^2 = 4(12) = 48$$
$$\text{Length of latus rectum } q = \frac{2b^2}{a} \Rightarrow q^2 = \frac{4b^4}{a^2} = \frac{4(144)}{8} = 72$$
$$p^2 + q^2 = 48 + 72 = 120$$
Ans. (120)