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`

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

The focus of the parabola y^(2)-4y-8x+4=0 is,

The focus of the parabola y^(2)-4y-8x-4=0 is

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

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 normals at three points P,Q,R of the parabola y^2=4ax meet in (h,k) The centroid of triangle PQR lies on (A) x=0 (B) y=0 (C) x=-a (D) y=a

A circle, with its centre at the focus of the parabola y^2 =4ax and touching its directrix, intersects the parabola at the point (A) (a, 2a) (B) (a,-2a) (C) (a/2, a) (D) (a/2, 2a)

If x-2y-a=0 is a chord of the parabola y^(2)=4ax , then its langth, is

The locus of a point on the variable parabola y^2=4a x , whose distance from the focus is always equal to k , is equal to ( a is parameter) (a) 4x^2+y^2-4k x=0 (b) x^2+y^2-4k x=0 (c) 2x^2+4y^2-9k x=0 (d) 4x^2-y^2+4k x=0

If the points (2,3) and (3,2) on a parabola are equidistant from the focus, then the slope of its tangent at vertex is (a) 1 (b) -1 (c) 0 (d) oo