if `y=4x+3` is parallel to a tangent to the parabola `y^2=12x`, then its distance from the normal parallel to the given line is

Options

A. `(213)/(sqrt(17))`
B. `(219)/(sqrt(17))`
C. `(211)/(sqrt(17))`
D. `(210)/(sqrt(17))`

Answer

B

Explanation

Given equation of parabola is `y^2=12x` ....(i)
On differenting both sides w.r.t. x, we get
`2y(dy)/(dx)=12rArr (dy)/(dx)=(6)/(y)`
Since , the noraml to the curve is parallel to the line `y=4x+3`
`therefore` Slope of normal curve = Slope of line
`rArr -(y)/(6)=4rArr y=-24`
From Eq.(i),`(-24)^2=12x`
`rArr 24xx24xx12x`
`rArr x=48`
`therefore` Normal point on a curve is `(48-24)`.
And distance from `(48,-24)` to the line `4x-y+3=0` is
`(4xx48+24+3)/(sqrt(4^2+1^2))=(219)/(sqrt(17))`