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question mark

  1. A ray of light incident at the point (-2,-1) gets reflected from the tangent at (0,-1) to the circle x^2 +y^2 =1 . The reflected ray touches the circle. The equation of the line along which the incident ray moved is ?? Options a) 4x -3y +11=0 b) 4x+3y+11=0 c) 3x +4y+11=0

Rahul bhardwaz , 7 Years ago
Grade 12th pass
anser 2 Answers
Arun
Let the equation of incident ray be 
y=m1x c1, 
and, reflected ray be, 
y=m2x c2, 

Equation of tangent to circle at(0,-1), 
is y=-1, slope 0, 
now this tangent passes thru (-2,-1), 
and the ray is incidnt at (-2,-1), 
Normal at this pt to the tangent is x=-2, slope infinity, 
now we know, angle of incidnc = angle of reflection, 
So, 
Lines y=m1x C1 
and y=m2x c2 
are equally inclind to normal at (-2,-1) x=-2!! 

Now using formula for slope,equally inclind lines. 

(m1-m)/1 m.m1 = (m-m2)/1 m.m2 

Where, m=slope of normal x=-2, which is infinity, 

Solving this, we get, m1 m2 =0! 

Now since the reflected ray, 
y=m2x c2, 
is a tangent to circle, 
using tangency conditon, 
(c2)^2 = 1 (m2)^2,.....(1) 
and also 
since it passes thru (-2,-1), 
c2=m2-2!.......(1) 

Solving(1) and (2) 
We get, m2=3/4, 

But we know m1 m2=0,(provd above) 

So, m1= -3/4; 

Now since 
y=m1x c1 also passes thru (-2,-1), 
Put in equation, we get 
c1=m1-2 
So, c1= -11/4, 

Now putting c1 and m1 in equation of incident ray, 
we get, the equation as, 
4y 3x 11 =
Last Activity: 7 Years ago
kkbisht
The answer is  option (b) 4x+3y+11=0
Let y=mx+c be the reflected ray  touching (tangent) the circle .Using the condition that this reflected ray is tangent if the pependicular drawn from the centre(0,0) of the unit Circle x2 + y2=1 we get
c/\sqrt{}(1+m2) = 1 => c=\sqrt{}(1+m2) therefore the equatio is y=mx+\sqrt{}(1+m2). As it passes thogt the point of incidence(-2,-1) 
 
we have     -1=m(-2) +\sqrt{}(1+m2). =>(2m-1)2=1+m2 =>4m2 +1-4m=1+m2 => 3m2-4m=0 => m=0 or m=4/3.
Now just see If this tangent makes an angle \alpha } with the normal at the point of incidence.then this tangent will make an angle 90-\alpha } with tanget at the point(0,-1) which is parrale to x-axis.
Then by definition slope  tan(90-\alpha })=4/3 or cot \alpha }= 4/3.
Now the incident ray makes an angle 90+\alpha } with the x-axis therefore its slope is tan(90+\alpha })=  -cot\alpha }= -4/3 ( from above)
Hence the equation of the incident ray is y=mx+c passing through (-2,-1) and slope m= -4/3
we get after simple calculation the equation as 4x+3y+11=0
 
kkbisht
Last Activity: 7 Years ago
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