Find the length of the side of an equilateral traingle inscribed in the parabola `y^2=4ax`, so that one angular point is at the vertex.

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

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

Area of an equilateral triangle is 17300 cm^2 . By taking each vertex of the triangle as centre, a circle is drawn at each vertex. The radius of thecircleis half of the length of the side of an equilateral triangle. Find the area of that portion of the triangle which is not included in the circle. (pi=3.14 and sqrt3=1.73 )

The vertex of the parabola y^2=4a(x+a) is:

Find the area of the equilateral triangle that can be inscribed in the circle : x^2 + y^2- 4x + 6y -3=0 .

An equilateral triangle is inscribed in the parabola y^2 = 4 ax , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

If ‘a’ is the length of one of the sides of an equilateral triangle ABC, baseBC lies on x-axis and vertex B is at the origin, find the coordinates,of the vertices of the triangle ABC.

The locus of the middle points of normal chords of the parabola y^2 = 4ax is-

A point P moves such that the sum of the slopes of the normals drawn from it to the hyperbola xy=4 is equal to the sum of the ordinates of feet of normals. The locus of P is a curve C. Q. The area of the equilateral triangle inscribed in the curve C having one vertex as the vertex of curve C is