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Conic Sections
Hyperbola
Hard

Question

Let the sum of the focal distances of the point P(4,3)\mathrm{P}(4,3) on the hyperbola H:x2a2−y2 b2=1\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1 be 8538 \sqrt{\frac{5}{3}}. If for H , the length of the latus rectum is ll and the product of the focal distances of the point P is m , then 9l2+6 m9 l^2+6 \mathrm{~m} is equal to :

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Solution

1. Key Concept: Focal Distances of a Hyperbola

For a standard hyperbola with the equation x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the foci are located at F1(−ae,0)F_1(-ae, 0) and F2(ae,0)F_2(ae, 0), where ee is the eccentricity. For any point P(x,y)P(x,y) on the hyperbola, its distances from the foci are given by: PF1=∣ex+a∣andPF2=∣ex−a∣PF_1 = |ex + a| \quad \text{and} \quad PF_2 = |ex - a| Since the given point P(4,3)P(4,3) has a positive x-coordinate (x=4x=4), it

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