Given the equation 3x^2 - 42x + 6y + 147 = 0 determine:
a) The Vertex V
b) If the parabola opens up, down, left, or 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)
The given equation is `3x^2 - 42x + 6y + 147 = 0`
a) Comparison with the standard form of equation (i) reveals that it is a parabola with vertical axis and having its vertex V at `(7, 0)` .
b) As its axis is vertical, it can either open up or open down. Since `p=-1/2 ` (i.e. p<0), it should open downwards.
c) The coordinates of the focus F of a parabola, having vertical axis is given by `( h, k+p)` .
Thus, the coordinates of the focus F of the given parabola is at `(7, 0+(-1/2))` , i.e `(7, -1/2)` . It is in fact half unit downward from the vertex.
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