The correct option is
B x−7y+26=0Let 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
⇒x−2y+6=0
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1708033/original_image_%285%29.png)
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+21−2=±−2−m1−2m
3(−1)=−2−m1−2m or
3(−1)=−(−2−m1−2m)
⇒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
y−4=17(x−2)
⇒7y−28=x−2
⇒x−7y+26=0
Hence, the correct option is
(a).