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A particle moves on the parabola y^2 = 4...

A particle moves on the parabola `y^2 = 4ax`. Its distance from the focus is minimum for the following values of `x` (A) `-1` (B) 0 (C) 1 (D) `a`

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A particle moves on y^(2)=4ax its distance from focus is minimum for following values of x(a)-1(b)0(c)+1(d)a

The focus of the parabola y^2= 4ax is :

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Prove that the point on the parabola y^(2) = 4ax (a gt0) nearest to the focus is its vertex.

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Focus of parabola y = ax^(2) + bx + c is

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The coordinates of a point on the parabola y^2=8x whose distance from the circle x^2+(y+6)^2=1 is minimum is (2,4) (b) (2,-4) (18 ,-12) (d) (8,8)

The coordinates of a point on the parabola y^2=8x whose distance from the circle x^2+(y+6)^2=1 is minimum is (2,4) (b) (2,-4) (18 ,-12) (d) (8,8)