Show that a parallel bundle of light rays parallel to the x-axis and incident on a parabolis reflecting surface given by `x=2by^(2)`, will pass through a single point called focus of the reflecting surface. Also, find the focal length.

Answer

A, B

Explanation

Let the ray is incident at a point `P=(x_(1), y_(1))` on the mirror. Then, slope at P,

`tan(alpha) =((dy)/(dt))_((x_(1)), y_(1)) =(1)/(4b(y_(1)))` ...(i)
`(alpha)=90^(@)-(theta)`
and`(beta)=2(theta)`
Now, the reflected ray is passing through `P(x_(1), y_(1))` and has a slope `tan(beta)`.Hence, the equation will be
`((y-y_(1)))/((x-x_(1)))=-(tan beta)`
`= (-tan 2 theta)=(-2 tan theta)/(1-tan^(2) theta)`
`:. (y-y_(1))=(2 cot (alpha))/(1+cot^(2)(alpha)) (x_(1)-(x))` ...(ii)
Further, `(x_(1))=4 (by_(1))^(2)` ...(III)
At `F, x=0` ...(iv)
From eq,(i) to eq.(iv),
we get `x=1/(8b)`
It shows that the coordintes of F are unique `((1)/(8b),0)`.
Hence, the reflected ray passing through one focus and the focal length is`(1)/(8b)` Hence proved.