# 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

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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`

`rArr 2(y^2-8y-4x-8)=0`

`rArr y^2-8y-4x-8=0`

`rArr (y^2-2*4*y+16)-4x-8-16=0`

`rArr (y-4)^2=4x+24`

`rArr (y-4)^2=4(x+6)`

This can also be written as:

`rArr (y-4)^2=4*1*(x+6)`

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.

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