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Let P be the point on the parabola y^(2)...

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

A

`SP=2sqrt(5)`

B

`SQ:QP=(sqrt(5)+1):2`

C

the x-intercept of the normal to the parabola at P is 2

D

the slope of the tangent to the circle at Q is `1/2`

Text Solution

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The correct Answer is:
A, D
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