one thing to notice is that, the directrix is the horizontal heat y = 4, meaning is a upright parabola, and thus the squared change is the "x". The directrix is listed below the focus point, an interpretation the parabola is opened upwards, that also way that the "p" street is positive, in this situation p = 1.
You are watching: What is the equation of the quadratic graph with a focus of (3, 6) and a directrix of y = 4?
The crest is always halfway between the focus allude and the directrix, in this case is at (3,5).
\rule34em0.25pt" alt="\bf \textitparabola vertex kind with focus suggest distance \\\\ \beginarrayllll 4p(x h)=(y k)^2 \\\\ \stackrel\textitwe"ll usage this one4p(y k)=(x h)^2 \endarray \qquad \beginarrayllll vertex\ ( h, k)\\\\ p=\textitdistance indigenous vertex come \\ \qquad \textit emphasis or directrix \endarray \\\\<0.35em> \rule34em0.25pt" align="absmiddle" class="latexformula"> ~\dotfill\\\\ ~\hfill y=\cfrac14(x3)^2+5~\hfill" alt="\bf \begincases h=3\\ k=5\\ p=1 \endcases\implies 4(1)(y5)=(x3)^2\implies y5=\cfrac14(x3)^2 \\\\<0.35em> ~\dotfill\\\\ ~\hfill y=\cfrac14(x3)^2+5~\hfill" align="absmiddle" class="latexformula">sdrta.net:
Vertex is in ~
(
3
,
−
1
)
, focus is at
(
3
,
−
15
16
)
and also
directrix is
y
=
−
1
1
16
.
Explanation:
y
=
4
(
x
−
3
)
2
−
1
Comparing with standard kind of vertex form equation
y
=
a
(
x
−
h
)
2
+
k
;
(
h
,
k
)
being vertex , we find here
h
=
3
,
k
=
−
1
,
a
=
4
. Therefore vertex is at
(
3
,
−
1
)
.
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Vertex is in ~ equidistance native focus and also directrix and also at opposite
sides . The distance of vertex indigenous directrix is
d
=
1
4

a

∴
d
=
1
4
⋅
4
=
1
16
. Due to the fact that
a
>
0
, the parabola opens up upwards and
directrix is below vertex. For this reason directrix is
y
=
(
−
1
−
1
16
)
=
−
17
16
=
−
1
1
16
and focus is at
(
3
,
(
−
1
+
1
16
)
)
or
(
3
,
−
15
16
)
Stepbystep explanation:
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