What is the equation in standard form of the parabola with a focus at (-18,30) and a directrix of y= 22?

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Answer 1 · 2017-11-12T11:58:37

The equation of parabola in standard form is
$(x+18)^2 = 16(y-26)$

Focus is at $(-18,30)$ and directrix is $y=22$ . Vertex is at midway

between focus and directrix. Therefore vertex is at

$(-18,(30+22)/2)$ i.e at $(-18, 26)$ . The vertex form of equation

of parabola is $y=a(x-h)^2+k ; (h.k) ;$ being vertex. Here

$h= -18 and k =26$ . So the equation of parabola is

$y=a(x+18)^2 +26$ . Distance of vertex from directrix is

$d= 26-22=4$ , we know $d = 1/(4|a|)$

$:. 4 = 1/(4|a|) or |a|= 1/(4*4)=1/16$ . Here the directrix is below

the vertex , so parabola opens upward and $a$ is positive.

$:. a=1/16$ . The equation of parabola is $y=1/16(x+18)^2 +26$

or $1/16(x+18)^2 = y-26 or (x+18)^2 = 16(y-26)$ or

$(x+18)^2 = 4*4(y-26)$ .The standard form is

$(x - h)^2 = 4p (y - k)$ , where the focus is $(h, k + p)$

and the directrix is $y = k - p$ . Hence the equation

of parabola in standard form is $(x+18)^2 = 16(y-26)$

graph{1/16(x+18)^2+26 [-160, 160, -80, 80]}