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  • A ray emanating from the point (5,0) is incident on the hyperbola <img alt="\mathrm{9 x^2-16 y^2=144}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B9%20x%5E2-16%20y%5E2%3D144%7D

A ray emanating from the point (5,0) is incident on the hyperbola \mathrm{9 x^2-16 y^2=144} at the point \mathrm{P(8,8 \sqrt{3})}. Find the equation of the reflected ray after first reflection.
 

Option: 1

\mathrm{ y=8 \sqrt{3}}


Option: 2

\mathrm{x=8}


Option: 3

\mathrm{ 13 y-8 \sqrt{3} x+40 \sqrt{3}=0}


Option: 4

\mathrm{8 \sqrt{3} x-13 y+40 \sqrt{3}=0}


Answers (1)

best_answer

Given hyperbola is \mathrm{\frac{x^2}{16}-\frac{y^2}{9}=1}
Here a=4 and b=3
So, foci are \mathrm{\left( \pm \sqrt{a^2+b^2}, 0\right) \equiv( \pm 5,0)}

Incident ray through \mathrm{F_1(5,0)} strikes the ellipse at point \mathrm{P(8,8 \sqrt{3})}. Therefore, reflected ray will go through another focus \mathrm{F_2(-5,0)}. So, reflected ray is line through points \mathrm{F_2} and P, which is \mathrm{8 \sqrt{3 x}-13 y+40 \sqrt{3}=0}

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rishi.raj

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