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determine the focus, vertex and directrix of the parabola of x^2+6x-8y+17=0

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determine  the focus, vertex and directrix of the parabola of x^2+6x-8y+17=0.

asked Feb 21, 2014 in GEOMETRY by andrew Scholar

1 Answer

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Given equation x ² + 6x  - 8y + 17 = 0

We first put the equation in to the form for a translated parabola (x - h )² = 4p (y - k)

To do this we complete the squre on the x  terms and move the other terms to right.

x ² + 6x = 8y - 17

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x coefficient is 6,(half the x coefficient)² = 9

x ² + 6x + 9 = 8y - 17 + 9

(x + 3)² = 8y - 8

In the next step we factored 8 from the right hand side to make the coefficient of our y  "+1", as this is the standard form.

(x + 3)² = 8(y - 1)

(x -( -3))² = 4(2)(y - 1)

Compare it to  parabola equation is (x - h )² = 4p (y - k ),

where (h , k ) = vertex and p = directed distance from vertex to focus.

p = 2

Vertex of parabola  = (-3, 1)

Focus = (h , k + p ) = [-3, 1+ 2)] = (-3, 3)

Equation of directrix is = k - p

y  = 1 - 2

y  = - 1

Vertex (-3, 1), focus (-3, 3) and directrix  y = - 1.

 

answered Apr 2, 2014 by david Expert

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