How do you find the vertex, focus, and directrix of the parabola $4 x - y^{2} - 2 y - 33 = 0$ $4x-y^2-2y-33=0$ ?

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Answer 1 · 2017-01-18T22:56:23

Please see the explanation.

Write the given equation in $x (y) = a y^{2} + b x + c$ $x(y) = ay^2 + bx + c$ form.

$4 x - y^{2} - 2 y - 33 = 0$ $4x - y^2 - 2y - 33 = 0$

$4 x = y^{2} + 2 y + 33$ $4x = y^2 + 2y + 33$

$x (y) = \frac{1}{4} y^{2} + \frac{1}{2} y + \frac{33}{4}$ $x(y) = 1/4y^2 + 1/2y + 33/4$

The y coordinate of the vertex, $k = - \frac{b}{2 a}$ $k = -b/(2a)$ :

$k = - \frac{\frac{1}{2}}{2 (\frac{1}{4})}$ $k = -(1/2)/(2(1/4))$

$k = - 1$ $k = -1$

The x coordinate of the vertex, $h = x (k)$ $h = x(k)$ :

$h = \frac{1}{4} {(- 1)}^{2} + \frac{1}{2} (- 1) + \frac{33}{4}$ $h = 1/4(-1)^2 + 1/2(-1) + 33/4$

$h = 8$ $h = 8$

The vertex is the point $(8, - 1)$ $(8, -1)$

The focal distance is, $f = \frac{1}{4 (a)}$ $f = 1/(4(a))$

$f = \frac{1}{4 (\frac{1}{4})}$ $f = 1/(4(1/4))$

$f = 1$ $f = 1$

The focus is located at the point $(h + f, k)$ $(h + f, k)$

$(8 + 1, - 1)$ $(8 + 1, -1)$

The focus is the point $(9, - 1)$ $(9, -1)$

The equation of the directrix is $x = h - f$ $x = h - f$

$x = 8 - 1$ $x = 8 - 1$

$x = 7$ $x = 7$

How do you find the vertex, focus, and directrix of the parabola 4x-y^2-2y-33=04x−y2−2y−33=04x-y^2-2y-33=0?