Find the directrix of a parabola
- Thread starter norton
- Start date
Hello, norton!
You are expected to know that the vertex is at \(\displaystyle (-3,\,7)\)
. . and the parabola opens to to the right: \(\displaystyle \subset\)
From \(\displaystyle p\,=\,1\), we know that the focus is one unit "inside" the parabola,
. . to the right of the vertex at \(\displaystyle (-2,7)\)
and know that the directrix is one unit "outside" the parabola,
. . to the left of the vertex: the vertical line: \(\displaystyle x\,=\,-4\)
Find the directrix of the parabola: \(\displaystyle \y\,-\,7)^2\:=\:4(x\,+\,3)\)
I took \(\displaystyle 4p\,=\,4\) to get \(\displaystyle p\,=\,1\;\;\) Correct!
Not sure where to go or if I did it correctly.
My choices for answers are:
\(\displaystyle (a)\;x\,=\,-4\;\;\;(b)\;x\,=\,-2\;\;\;(c)\;x\,=\,2\;\;\;(d)\;x\,=\,6\)
You are expected to know that the vertex is at \(\displaystyle (-3,\,7)\)
. . and the parabola opens to to the right: \(\displaystyle \subset\)
From \(\displaystyle p\,=\,1\), we know that the focus is one unit "inside" the parabola,
. . to the right of the vertex at \(\displaystyle (-2,7)\)
and know that the directrix is one unit "outside" the parabola,
. . to the left of the vertex: the vertical line: \(\displaystyle x\,=\,-4\)