Question #a3b13

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Answer 1 · 2017-03-20T19:53:05

Please see the explanation.

Given: #y^2-8x+2y+25=0#

Write in standard form, #x = ay^2+by +c#:

#x = 1/8y^2+2/8y+25/8#

#a = 1/8, b = 2/8, and c = 25/8#

The y coordinate, k, of the vertex and the focus is:

#k = -b/(2a)#

#k = (-2/8)/(2(1/8))#

#k = -1#

The x coordinate, h, of the vertex is the equation evaluated at y = k= -1:

#h = 1/8(-1)^2+2/8(-1)+25/8#

#h = 3#

The vertex is the point #(3,-1)#

Find the focal distance, f:

#f = 1/(4a)#

#f = 1/(4(1/8))#

#f = 2#

The x coordinate of the focus is the x coordinate of the vertex plus the focal distance and the y coordinate is the same.

The focus is #(5, -1)#

The directrix is a vertical line at the x coordinate of the vertex minus the focal distance:

#x = 3-2#

#x = 1# is the equation of the directrix.

For the x intercept evaluate the equation at #y = 0#:

#x = 25/8#

Evaluate the discriminant:

#b^2-4(a)(c) = (2/8)^2-(4)(1/8)(25/8) = 4/64-100/64 = -96/64#

There are no real y intercepts.