What is the standard form equation of the parabola with a directrix of x=5 and focus at (11, -7)?

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What is the standard form equation of the parabola with a directrix of x=5 and focus at (11, -7)?

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Answer 1 · 2017-05-03T03:50:10

Standard form is:

$x = \frac{1}{12} y^{2} + \frac{14}{12} y + \frac{145}{12}$ $x = 1/12y^2+14/12y+ 145/12$

Explanation:

Because the directrix is a vertical line, $x = 5$ $x = 5$ , the vertex form for the equation of the parabola is:

$x = \frac{1}{4 f} {(y - k)}^{2} + h [1]$ $x = 1/(4f)(y-k)^2+h" [1]"$

where (h,k) is the vertex and #f is the signed horizontal distance from the vertex to the focus.

We know that the y coordinate, k, of the vertex is the same as the y coordinate of the focus:

$k = - 7$ $k = -7$

Substitute -7 for k into equation [1]:

$x = \frac{1}{4 f} {(y - - 7)}^{2} + h [2]$ $x = 1/(4f)(y--7)^2+h" [2]"$

We know that the x coordinate of the vertex is the midpoint between the x coordinate of the focus and the x coordinate of the directrix:

$h = \frac{x_{focus} + x_{directrix}}{2}$ $h= (x_"focus"+x_"directrix")/2$

$h = \frac{11 + 5}{2}$ $h= (11+5)/2$

$h = \frac{16}{2}$ $h = 16/2$

$h = 8$ $h = 8$

Substitute 8 for h into equation [2]:

$x = \frac{1}{4 f} {(y - - 7)}^{2} + 8 [3]$ $x = 1/(4f)(y--7)^2+8" [3]"$

The focal distance is the signed horizontal distance from the vertex to the focus:

$f = x_{focus} - h$ $f = x_"focus"-h$

$f = 11 - 8$ $f = 11-8$

$f = 3$ $f = 3$

Substitute 3 for f into equation [3]:

$x = \frac{1}{4 (3)} {(y - - 7)}^{2} + 8$ $x = 1/(4(3))(y--7)^2+8$

We will multiply the denominator and write -- as +

$x = \frac{1}{12} {(y + 7)}^{2} + 8$ $x = 1/12(y+7)^2+8$

Expand the square:

$x = \frac{1}{12} (y^{2} + 14 y + 49) + 8$ $x = 1/12(y^2+14y+ 49)+8$

Distribute the $\frac{1}{12}$ $1/12$

$x = \frac{1}{12} y^{2} + \frac{14}{12} y + \frac{49}{12} + 8$ $x = 1/12y^2+14/12y+ 49/12+8$

Combine the constant terms:

$x = \frac{1}{12} y^{2} + \frac{14}{12} y + \frac{145}{12}$ $x = 1/12y^2+14/12y+ 145/12$

Answer 2 · 2017-05-03T06:39:53

$x = \frac{y^{2}}{12} + \frac{7}{6} y + \frac{145}{12}$ $x=y^2/12+7/6y+145/12$

Explanation:

Directrix $x = 5$ $x=5$
Focus $(11, - 7)$ $(11, -7)$
From this we can findout the vertex.
Look at the diagram
enter image source here

Vertex lies exactly in between Directrix and Focus

$x, y = \frac{5 + 11}{2}, \frac{- 7 + (- 7)}{2} = (8, - 7)$ $x,y=(5+11)/2, (-7 + (-7))/2=(8, -7)$

The distance between Focus and vertex is $a = 3$ $a=3$
The parabola is opening to the right
The equation of the Parabola here is -

${(y - k)}^{2} = 4 a (x - h)$ $(y-k)^2=4a(x-h)$
$(h, k)$ $(h,k)$ is the vertex
$h = 8$ $h=8$
$k = - 7$ $k=-7$

Plugin $h = 8; k = - 7 and a = 3$ $h=8; k=-7 and a=3$ in the equation

${(y - (- 7))}^{2} = 4.3 (x - 8)$ $(y-(-7))^2=4.3(x-8)$
${(y + 7)}^{2} = 4.3 (x - 8)$ $(y+7)^2=4.3(x-8)$

$12 x - 96 = y^{2} + 14 y + 49$ $12x-96=y^2+14y+49$ [by transpose]
$12 x = y^{2} + 14 y + 49 + 96$ $12x=y^2+14y+49+96$
$12 x = y^{2} + 14 y + 145$ $12x=y^2+14y+145$
$x = \frac{y^{2}}{12} + \frac{14}{12} y + \frac{145}{12}$ $x=y^2/12+14/12y+145/12$
$x = \frac{y^{2}}{12} + \frac{7}{6} y + \frac{145}{12}$ $x=y^2/12+7/6y+145/12$

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What is the standard form equation of the parabola with a directrix of x=5 and focus at (11, -7)?

2 Answers

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