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=2sqrt5`
B. `SP:QP=sqrt5+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

B

Explanation

Let the coordiantes of P eb `(t^(2), 2t)`. The coordinates of the centre S of the circle are (2, 8). If P is at the shortest distance from the centre s of the circle. Then, the normal to the parabola at P must pass through the centre S of the circle. The equation of the normal to the parabola at `P(t^(2), 2t)` is
`y+tx=2t+t^(3)" ...(i)"`

If it passes through S(2, 8). then
`8+2t=2t+t^(3)rArrt^(3)=8rArrt=2`
So, the coordinates of are P(4, 4)
`:." "SP=sqrt((2-4)^(2)+(8-4)^(2))=2sqrt5`,
`SQ="radius"=sqrt(4+64-64)=2`
`PQ=SP-SQ=2sqrt(5)-2=2(sqrt(5)-1)` ,
`:. (SQ)/(PQ)=(2)/(2(sqrt(5)-1))=(sqrt(5)+1)/(4)`