Let P be the point on parabola `y^2=4x` which is at the shortest distance from the center S of the circle `x^2+y^2-4x-16y+64=0` let Q be the point on the circle dividing the line segment SP internally. Then

Options

A. `SP = 2 sqrt(5)`
B. `SQ : QP (sqrt(5) + 1) : 2`
C. the x-intercept of the normal to the parabola at P is 6
D. the slope of the tangent to the circle at Q is `(1)/(2)`

Answer

A, C, D

Explanation

Tangent to `y^(2) = 4x` at `(t^(2), 2t)` is

`y(2 t) = 2(x + t^(2))`
`implies yt = x + t^(2)`
Equation of normal of `P(t^(2) , 2t)` is
`y + tx = 2t + t^(3)`
Since, normal at P passes through centre of circle S(2,8).
`:. 8 + 2 t + t^(3)`
`implies t = 2,` i., P (4,4)
[Since, shortest distance between two curves lie along their common normal and the common normal will pass through the centre of circle]
`:. SP = sqrt((4 - 2)^(2) + (4 - 8)^(2)) = 2 sqrt(5)`
`:.` Opion (a) is correct.
Also, `SQ = 2`
`:. PQ = SP - SQ = 2 sqrt(5) - 2`
Thus, `(SQ)/(QP) = (1)/(sqrt(5) - 1) = (sqrt(5) + 1)/(4)`
`:.` Option (b) is wrong
Now x intercept of normal is `x = 2 + 2^(2) = 6`
`:.` Option (c ) is corect
Slope of tangent `= (1)/(t) = (1)/(2)`
`:.` Option (d) is correct.