How do I find the directrix of the parabola whose equation is $y = \frac{x^{2}}{32}$ $y=x^2/32$ ?

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How do I find the directrix of the parabola whose equation is $y = \frac{x^{2}}{32}$ $y=x^2/32$ ?

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Answer 1 · 2014-10-19T05:12:52

You have a parabola with vertex at $(0, 0)$ $(0, 0)$ that is opening upward.

That means your directrix is horizontal.

To get the directrix, transfer $x^{2}$ $x^2$ 's coefficient to the other side.
Equate the resulting coefficient of $y$ $y$ to $4 d$ $4d$ , where $d$ $d$ is the distance from the vertex to the directrix

$y = \frac{x^{2}}{32}$ $y = x^2/32$
$\Rightarrow 32 y = x^{2}$ $=> 32y = x^2$

$4 d = 32$ $4d = 32$
$\Rightarrow d = 8$ $=> d = 8$

Since the parabola is opening upward, the directrix is below the vertex.

$y = h - d$ $y = h - d$

$y = 0 - 8$ $y = 0 - 8$

$y = - 8$ $y = -8$

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How do I find the directrix of the parabola whose equation is $y = \frac{x^{2}}{32}$ $y=x^2/32$ ?

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