The mirror image of the parabola `y^2=4x` in the tangent to the parabola at the point (1, 2) is `(x-1)^2=4(y+1)` (b) `(x+1)^2=4(y+1)` `(x+1)^2=4(y-1)` (d) `(x-1)^2=4(y-1)`

Options

A. `(x-1)^(2)=4(y+1)`
B. `(x+1)^(2)=4(y+1)`
C. `(x+1)^(2)=4(y-1)`
D. `(x-1)^(2)=4(y-1)`

Answer

C

Explanation

(3) Any point on the given parabola is `(t^(2),2t)`
The equation of the tangents at (1,2) is x-y+1=0.
The image (h,k) of the point `(t^(2),2t)` on x-y+1=0 is given by
`(h-t^(2))/(1)=(k-2t)/(-1)=-(2(t^(2)-2t+1))/(1+1)`
`:." "h=t^(2)+t^(2)+2t-1=2t-1`
`andk=2t+t^(2)-2t+1=t^(2)+1`
Eliminating t from `h=2t-1andk=t^(2)+1`, we get
`(h+1)^(2)=4(k-1)`
The required equation of reflection is `(x+1)^(2)=4(y-1)`.