We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.

The equation of the parabola is either of the from x^{2 }= 4ay or x^{2} = -4ay.

Given that the parabola passes through point (5, 2), which lies in the first quadrant.

So, the equation of the parabola is of the form

x^{2} = 4ay, while point (5, 2) must satisfy the equation x^{2} = 4ay.

Then,

5^{2} = 4a(2)

25 = 8a

a = \( \dfrac{25}{8}\)

Thus, the equation of the parabola is

x^{2} = 4\(
(\dfrac{25}{8})y\)

x^{2} = \(
\dfrac{25y}{2}\)

2x^{2} = 25y

The equation of the parabola is 2x^{2} = 25y

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

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.