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.