##### Given the followinng parabola

label Algebra
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x^2-6x+8y-7=0

Step 1. Find the focus of the parabola

Step 2. Find the directrix of the parabola

Step 3. Ffind the vertex and the two points which lie on the line through the focus that is parallel to the directrix.

Oct 6th, 2015

The first part for solving this is rewriting this equation in vertex form y = a(x-h)^2 + k

8y = -x^2 +6x + 7

y = -(x^2-6x-7)/8

Then we have to complete the square:

y=-[(x-3)^2 - 9 - 7]/8 = -[(x-3)^2]/8 + 2

Going back to standard form, this says a = -1/8, h = 3 and k =2.

1. When the equation is in vertex form, the focus is at the point (h, k + 1/(4a) ). Replacing h, k, and a we get that the focus is at (3, 2-1/(4/8) ) = (3, 0)

Step 2. The directrix of the parabola is given by the line y = k -1/(4a) = 2 + 2 = 4

y = 4 is the equation of the directrix

Step 3

The vertex is located at (h,k), or (3, 2)

The points which lie on the line through the focus and parallel to the directrix are the points that have y = 0

0 = -(x^2-6x-7)/8

(x-7)(x+1) = 0

x = 7, -1

So the points are (7,0), (-1,0)

Please let me know if you need any clarification. Always glad to help!
Oct 6th, 2015

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Oct 6th, 2015
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Oct 6th, 2015
Oct 18th, 2017
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