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  • A ray  emanating from the point is incident on the hyperbola <img alt="\mathrm{9 x^2-16 y^2=144}"

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} with abscissa \mathrm{8} ; then the equation of reflected ray after first reflection is (Point \mathrm{P} lies in first quadrant)

Option: 1

\mathrm{3 \sqrt{3} x-13 y+15 \sqrt{3}=0}


Option: 2

\mathrm{3 x-13 y+15=0}


Option: 3

\mathrm{3 \sqrt{3} x+13 y-15 \sqrt{3}=0}


Option: 4

None of these


Answers (1)

best_answer

Given hyperbola is \mathrm{9 x^2-26 y^2=144}. This equation can be rewritten as \mathrm{\frac{x^2}{16}-\frac{y^2}{9}=1\ \ .............(i)}
Since \mathrm{x} coordinate of \mathrm{P} is \mathrm{8}. Let \mathrm{y}- coordinate of \mathrm{P} is \mathrm{\alpha }
\begin{aligned} &\mathrm{ \therefore \quad(8, \alpha) \text { lies on (i) }} \\ &\mathrm{ \therefore \quad \frac{64}{16}-\frac{\alpha^2}{9}=1 ; \quad \therefore \quad \alpha=27} \\ &\mathrm{ \alpha=3 \sqrt{3}} \end{aligned} (P lies in first quadrant)
Hence coordinate of point \mathrm{P} is \mathrm{(8,3 \sqrt{3})}
\mathrm{\mathbb{B}} Equation of reflected ray passing through \mathrm{P(8,3 \sqrt{3}) \text { and } S^{\prime}(-5,0) ;\ \ \ \therefore}  Its equation is \mathrm{y-3 \sqrt{3}=\frac{0-3 \sqrt{3}}{-5-8}(x-8)}
\mathrm{\text { or } 13 y-39 \sqrt{3}=3 \sqrt{3} x-24 \sqrt{3} \quad \text { or } 3 \sqrt{3} x-13 y+15 \sqrt{3}=0}

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Nehul

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