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`

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 :

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

The length of a focal chord of the parabola y^(2) = 4ax at a distance b from the vertex is c. Then

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

y= -2x+12a is a normal to the parabola y^(2)=4ax at the point whose distance from the directrix of the parabola is

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

If the length of a focal chord of the parabola y^(2)=4ax at a distance b from the vertex is c then prove that b^(2)c=4a^(3).

Let f(x)=4x^2-4ax+a^2-2a+2 be a quadratic polynomial in x, a be any real number, If x- coordinate of vertex of parabola y=f(x) is less than 0 and f(x) has minimum value 3 for x in [0,2], then value of a is (a) 1+sqrt(2) (b) 1-sqrt(2) (c) 1-sqrt(3) (d) 1+sqrt(3)