Vertex of `y=x^2-2ax+1` is closest to the origin then

The vertex of the parabola y=x^(2)-2ax+1 is closest to the origin when a =

The point on the line 3x-2y=1 which is closest to the origin is

Find the point on the line 2x+3y =6 which is closest to the origin.

Find the point on the straight line 2x+3y=6 , which is closest to the origin.

A square is inscribed in the circle x^(2)+y^(2)-6x+8y-103=0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is

If the curve y=ax^(2)+bx+c passes through the point (1, 2) and the line y = x touches it at the origin, then

If the curve y=ax^(2)+bx+c passes through the point (1, 2) and the line y = x touches it at the origin, then

An equilateral triangle is Inscribed in parabola y^(2) = 4ax whose one vertex is at origin then length of side of triangle

If (x_(1) ,y_(1))" and " (x_(2) , y_(2)) are ends of a chord of y^(2) = 4ax , which cuts its axis at a distance delta from the origin , then the product x_(1) x_(2) =

show that the equation of the parabola whose vertex and focus are on the x -axis at distances a and a' from the origin respectively is y^(2) = 4 (a'-a) (x - a )