SOLUTION: for x - 4 = ( y + 5 ) ^2 would (4,-5) be the vertex? and what is the focus and directrix? also, what is the focus and directrix of -3( y +3)^2 = - (x+4) and -3 (x+2)^2 =

SOLUTION: for x - 4 = ( y + 5 ) ^2 would (4,-5) be the vertex? and what is the focus and directrix? also, what is the focus and directrix of -3( y +3)^2 = - (x+4) and -3 (x+2)^2 = - (y+

Question 546364: for x - 4 = ( y + 5 ) ^2
would (4,-5) be the vertex?
and what is the focus and directrix?
also, what is the focus and directrix of -3( y +3)^2 = - (x+4) and -3 (x+2)^2 = - (y+6)
I was confused because I wasn't sure what the point was so I couldn't figure out the focus and directrix.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
for x - 4 = ( y + 5 ) ^2
would (4,-5) be the vertex?
and what is the focus and directrix?
also, what is the focus and directrix of -3( y +3)^2 = - (x+4) and -3 (x+2)^2 = - (y+6)
I was confused because I wasn't sure what the point was so I couldn't figure out the focus and directrix.
**
The following are 4 standard forms of equations for parabolas showing focus and directrix:
(axis of symmetry vertical or horizontal)
(x-h)^2=4p(y-k), (h,k)=(x,y) of vertex, parabola opens upward)
(x-h)^2=-4p(y-k), (h,k)=(x,y) of vertex, parabola opens downward)
(y-k)^2=4p(x-h), (h,k)=(x,y) of vertex, (parabola opens rightward)
(y-k)^2=-4p(x-h), (h,k)=(x,y) of vertex, (parabola opens leftward)
..
For given equation, x - 4 = ( y + 5 ) ^2, it could be rewritten: (y+5)^2=(x-4), which looks like the third form listed above. In this case, 4p=1, so p=1/4. The focus and directrix are p units from the vertex on the axis of symmetry. You are correct in saying the vertex is (4,-5).
axis of symmetry: y=-5
focus: (4+1/4,-5)=(17/4,-5)
directrix: x=4-1/4=15/4
..
For given equation, -3( y +3)^2 = - (x+4), it could be rewritten: (y+3)^2=(1/3)(x+4), which also looks like the third form listed above. In this case, however, 4p=1/3, so p=1/12. The focus and directrix are p units from the vertex on the axis of symmetry.
vertex: (-4,-3).
axis of symmetry: y=-3
focus: (-4+1/12,-3)=(-47/12,-3)
directrix: x=-4-1/12=-49/12
..
For given equation, -3 (x+2)^2 = - (y+6), it could be rewritten: (x+2)^2=(1/3)(y+6), which looks like the first form listed above. In this case, 4p=1/3, so p=1/12. The focus and directrix are p units from the vertex on the axis of symmetry.
vertex: (-2,-6).
axis of symmetry: x=-2
focus: (-2,-6+1/12)=(-2,-71/12)
directrix: y=-73/12

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