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Question

A ray of light is moving along the line y=x+2, gets reflected from a parabolic mirror whose equation is
y2=4(x+2). The reflection does not happen at the vertex of the parabola. After reflection, the ray does not pass through the focus of the parabola. What will be the equation of the line which contains the reflected ray?

A
x7y10=0
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B
x7y+26=0
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C
x+7y+10=0
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D
x+7y10=0
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Solution

The correct option is B x7y+26=0
Let the line y=x+2 intersect the parabola at P.
Solving the parabola and the line,
(x+2)2=4(x+2)
(x+2)=4
x=2
y=2+2=4.

Therefore, the point P is (2,4).
Now the equation of the tangent to the parabola
y2=4x+8 at P(2,4) is yy1=2(x+x1)+8
4y=2(x+2)+8
x2y+6=0


Let IP be the incident ray, PM be the reflected ray and PN be the normal. We know that the normal is equally inclined with the incident ray and the reflected ray.
We have the slope of IP=1, Slope of tangent at P=1/2. Thus, the slope of normal PN=2.
Let the slope of the reflected ray PM be m.
Then, angle between the incident ray IP and normal PN is equal to the angle between the normal PN and reflected ray PM.

Thus, using the formula of angle between two lines, we get
1+212=±2m12m
3(1)=2m12m or 3(1)=(2m12m)
m=17 or 1

Now, we cannot take m=1 as m=1 will give us the original incident ray itself.
Hence, the equation of the reflected ray is
y4=17(x2)
7y28=x2
x7y+26=0
Hence, the correct option is (a).

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