Given:

The equation is x^{2} = 6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with x^{2} = 4ay, we get,

4a = 6

a = \( \dfrac{6}{4}\)

= \( \dfrac{3}{2}\)

Thus, the co-ordinates of the focus = (0,a) = (0,\( \dfrac{3}{2}\))

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

The equation of directrix, y =-a, then,

y =\( \dfrac{-3}{2}\)

Length of latus rectum = 4a = 4(\( \dfrac{3}{2}\)) = 6

Answered by Aaryan | 1 year agoAn 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.