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A ball moving around the circle \mathrm{x^2+y^2-2 x-4 y-20=0} in anti-clockwise direction leaves it tangentially at the point \mathrm{\mathrm{P}(-2,-2)}. After getting reflected from a straight line, it passes through the center of the circle. Find the equation of the straight line if its perpendicular distance from \mathrm{\mathrm{P}\: is\: 5 / 2}. You can assume that the angle of incidence is equal to the angle of reflection.

Option: 1

\mathrm{(4 \sqrt{3}-3) x-(4+3 \sqrt{3}) y-(39-2 \sqrt{3})=0}
 


Option: 2

\mathrm{(4 \sqrt{3}+3) x-(4-3 \sqrt{3}) y-(39-2 \sqrt{3})=0}
 


Option: 3

\mathrm{(4 \sqrt{3}-3) x+(4+3 \sqrt{3}) y-(39-2 \sqrt{3})=0}
 


Option: 4

None of these


Answers (1)

best_answer

Radius of the circle =\mathrm{CP}=\sqrt{9+16}=5

Let the equation of surface is \mathrm{y=m x+c}

Given \mathrm{ P Q=5 / 2}

\mathrm{\therefore \quad \frac{-2 \mathrm{~m}+2+\mathrm{c}}{\sqrt{\left(1+\mathrm{m}^2\right)}}= \pm 5 / 2}             ............(1)

Tangent at \mathrm{P} strikes it at the point \mathrm{M} and after reflection passes through the center \mathrm{\mathrm{C}(1,2).}

Let \mathrm{MN} be the normal at \mathrm{\mathrm{M}}

\angle \mathrm{PMN}=\angle \mathrm{NMC}=\alpha

in \triangle \mathrm{PCM},

\mathrm{\tan 2 \alpha =\frac{P C}{P M} }
\mathrm{\Rightarrow \quad \tan 2 \alpha =\frac{5}{P M}}                       ............(2)

and in \mathrm{\triangle P Q M}

\mathrm{ \sin (90-\alpha)=\frac{5 / 2}{\mathrm{PM}} }
\mathrm{ \therefore \quad \mathrm{PM}=\frac{5}{2 \cos \alpha} }                       ..............(3)
from (2) & (3)
\mathrm{ 5 \cot 2 \alpha=\frac{5}{2 \cos \alpha} }

\mathrm{ \Rightarrow \quad 2 \cot 2 \alpha \cos \alpha=1 }

\mathrm{ \Rightarrow \quad \frac{2 \cos 2 \alpha}{\sin 2 \alpha} \cdot \cos \alpha=1 }

\mathrm{ \Rightarrow \quad \frac{2\left(1-2 \sin ^2 \alpha\right) \cos \alpha}{2 \sin \alpha \cos \alpha}=1 }

\mathrm{ \Rightarrow \quad 1-2 \sin ^2 \alpha=\sin \alpha }

\mathrm{ \Rightarrow \quad 2 \sin ^2 \alpha+\sin \alpha-1=0 }

\mathrm{ \Rightarrow \quad(2 \sin \alpha-1)(\sin \alpha+1)=0 }

\mathrm{ \Rightarrow \quad \sin \alpha \neq-1 \quad \therefore \sin \alpha=\frac{1}{2} }

\mathrm{ \therefore \quad \alpha=30^{\circ} }

Tangent at \mathrm{P(-2,-2)} is

\mathrm{ \Rightarrow \quad-2 x-2 y-(x-2)-2(y-2)-20=0 }

\mathrm{ \text { Slope of } \quad 3 \mathrm{x}+4 \mathrm{y}+14=0 }

\mathrm{ \because \quad \angle \mathrm{PMQ}=90-\alpha=90^{\circ}-30^{\circ}=60^{\circ} }

\mathrm{ \therefore \quad \tan 60^{\circ}=\left|\frac{\mathrm{m}+3 / 4}{1-3 \mathrm{~m} / 4}\right| }

\mathrm{ \qquad \sqrt{3}=\frac{4 \mathrm{~m}+3}{4-3 \mathrm{~m}} }

\mathrm{ \therefore \quad \mathrm{m}=\frac{4 \sqrt{3}-3}{4+3 \sqrt{3}} }

from (1) \mathrm{ \quad \pm \frac{5}{2}=\frac{2(1-\mathrm{m})+\mathrm{c}}{\sqrt{1+\mathrm{m}^2}} }

we get  \mathrm{\mathrm{c}=\frac{11+2 \sqrt{3}}{4+3 \sqrt{3}} \text { or } \frac{-39+2 \sqrt{3}}{4+3 \sqrt{3}}}

\mathrm{c} being intercept on \mathrm{y}-axis made by surface is clearly -ve.

Hence the required line is

\mathrm{ y=\left(\frac{4 \sqrt{3}-3}{4+3 \sqrt{3}}\right) x+\left(\frac{-39+2 \sqrt{3}}{4+3 \sqrt{3}}\right) }

\mathrm{(4 \sqrt{3}-3) x-(4+3 \sqrt{3}) y-(39-2 \sqrt{3})=0 .}

Hence option 1 is correct.
 







 

 

Posted by

rishi.raj

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