A ray emanating from the point (5, 0) is meident on the hyperbola `9x^2-16 y^2=144` at the point `P` with abscissa 8. Find the equation of the reflect

Question

A ray emanating from the point (5, 0) is meident on the hyperbola `9x^2-16 y^2=144` at the point `P` with abscissa 8. Find the equation of the reflect

A ray emanating from the point (5, 0) is meident on the hyperbola `9x^2-16 y^2=144` at the point `P` with abscissa 8. Find the equation of the reflected ray after the first reflection if point `P` lies in the first quadrant.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Given hyperbola is
`(x^(2))/(16)-(y^(2))/(9)=1" (1)"`
Here, a = 4 and b = 3.
So, foci and `(pmsqrt(a^(2)+b^(2)),0)-=(pm5,0).`
Incident ray through `F_(1)(5, 0)` strikes the ellipse at point `P(8, 3sqrt3)`.
Therefore, reflected ray will go through another focus `F_(2)(-5,0).`
So, reflected ray is line through points `F_(2) and P`, which is
`3sqrt3x-13y+15sqrt3=0.`