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Given the equation 3x^2 - 42x + 6y + 147 = 0 determine: a)  The Vertex V b)  If the...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 20, 2013 at 2:21 AM via web

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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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llltkl | College Teacher | Valedictorian

Posted July 20, 2013 at 3:11 AM (Answer #1)

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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`

`rArr x^2-14x+2y+49=0`

`rArr x^2-2*7x+49=-2y`

`rArr (x-7)^2=4*(-1/2)*(y-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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