A triangle has one vertex at `(0,0)` and the other two on the graph of `y=-2x^(2)+54` at `(x,y)` and `(-x,y)` where `0 < x < sqrt(27)` . The value of x so that the corresponding triangle has maximum area is

If the area of the triangle whose one vertex is at the vertex of the parabola, y^(2) + 4 (x - a^(2)) = 0 and the other two vertices are the points of intersection of the parabola and Y-axis, is 250 sq units, then a value of 'a' is

The vertex of the parabola y ^(2) -4y-x+3=0 is

A triangle is formed by vertices (0, 0), (x, y), (-x, y) on xy-plane. If the point (x, y) and (-x, y) lies on y =-x^2 54 , then maximum area of triangle is

The vertex of the parabola y^(2)+6x-2y+13=0 is

Find the equation of the line passing through the origin and dividing the curvilinear triangle with vertex at the origin, bounded by the curves y = 2x – x^(2) , y = 0 and x = 1 into two parts of equal area.

Vertex of the parabola x ^(2) +4x + 2y -7=0 is

The extremities of the base of an isosceles triangle are (2,0) and (0,2). If the equation of one of the equal side is x=2, then the equation of other equal side is x+y=2 (b) x-y+2=0y=2 (d) 2x+y=2

A triangle has vertices A_(i) (x_(i),y_(i)) for i= 1,2,3,. If the orthocenter of triangle is (0,0) then prove that |{:(x_(2)-x_(3),,y_(2)-y_(3),,y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))),(x_(3)-x_(1) ,,y_(3)-y_(1),,y_(2)(y_(3)-y_(1))+x_(2)(x_(3)-x_(1))),( x_(1)-x_(2),,y_(1)-y_(2),,y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))):}|=0

Vertex of the parabola y ^(2) + 2y + x=0 lies in the