Given the equation 2y^2 - 8x - 16y - 16 = 0 determine: a) The Vertex V b) If the parabola opens up, down, left, right c) The Focus F
The standard form of the equation of a parabola with vertex at (h, k) is given by:
`(x-h)^2=4p(y-k)` (where the axis is vertical) ---- (i)
and `(y-k)^2=4p(x-h)` (where the axis is horizontal) --- (ii)
Given equation is `2y^2 - 8x - 16y - 16 = 0`
This can also be written as:
a) Comparison with the standard form of equation (ii) reveals that it is a parabola with horizontal axis and having its vertex V at (-6, 4).
b) As its axis is horizontal, it can either open left or open right. Since p=1 (i.e. p>0), it should open towards right.
c) The coordinates of the focus F of a parabola, having vertical axis is given by ( h+p, k).
Thus, the coordinates of the focus F = (-6+1, 4), i.e (-5, 4). It is in fact one unit to the right of the vertex.