vertex of the parabola `9x^2-6x + 36y + 9 = 0` is

Find the area of the triangle formed by the lines joining the vertex of the parabola x^2 = - 36y to the ends of the latus rectum.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^2 = - 36y to the ends of the latus rectum.

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

The vertex of the parabola x^(2)-6x+4y+1=0 is

Find the area of the triangle found by the lines joining the vertex of the parabola x^(2) -36y to the ends of the latus rectum.

The vertex of the parabola x^(2)+12 x-9 y=0 is

Let l_(1) be the length of the latus rectum of the parabola 9x^(2)-6x+36y+19=0. Let l_(2), be the length of the latus rectum of the hyperbola 9x2-16y^(2)-18x-32y-151=0. Let n be the unit digit in the product of l_(1) and l_(2) The number of terms in the binomial expansion of (1+x)^(n) is

The distance between the vertex of the parabola y = x^(2) - 4x + 3 and the centre of the circle x^(2) = 9 - (y - 3)^(2) is

The distance between the vertex of the parabola y = x^2 - 4x + 3 and the centre of the circle x^2 = 9 - (y- 3)^2 is.