7. What is the equation of the parabola, in vertex form, with vertex at (-2,-4) and directrix x = -6?
(Points : 2)
Options:
(x+4)^2=8(y+2)
(y+4)^2=--8(x+6)
(x+2)^2=8(y+4)
(y+4)^2=16(x+2) <----- I think it is this one.
(Points : 2)
Options:
(x+4)^2=8(y+2)
(y+4)^2=--8(x+6)
(x+2)^2=8(y+4)
(y+4)^2=16(x+2) <----- I think it is this one.
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You think right.
Solution:
x - h = 1/4p*(y - k)^2
(x + 2) = 1/4p*(y + 4)^2
The parabola opens to the right. The vertex is midway between the focus and directrix. The focus is at (2 , -4) , then p = 4 => 4p =16
The equation is:
(x + 2) = 1/16*(y + 4)^2
or:
(y+4)^2=16(x+2)
Edit: your 2nd question is False: focus ia at(1 , 0)
Solution:
x - h = 1/4p*(y - k)^2
(x + 2) = 1/4p*(y + 4)^2
The parabola opens to the right. The vertex is midway between the focus and directrix. The focus is at (2 , -4) , then p = 4 => 4p =16
The equation is:
(x + 2) = 1/16*(y + 4)^2
or:
(y+4)^2=16(x+2)
Edit: your 2nd question is False: focus ia at(1 , 0)
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You are welcome my friend, and I'm glad it worked out OK for you.
Best Regards.
Best Regards.
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The vertex isn't at the centre, so it's a translated parabola, with vertex (or centre if you want) at (h,k)
The general formula(s) for translated parabolas:
(y-k)^2=4a(x-h), with focus (h+a,k), and directrix at x=h-a.
OR
(x-h)^2=4a(y-k), focus (h,k+a) and directrix y=k-a
(so the vertex is (-2,-4), so h=-2 and k=-4.)
Since the directrix is "x" equals -6, it tells us that we should use the first formula for a translated parabola (shown above).
Sub (h,k) points into the equation
---> (y--4)^2=4a(x--2)
--->(y+4)^2=4a(x+2)
To find a, use the equation when finding a directrix (x=h-a)
x is given, (-6) and we know h.
-6=-2-a
a=4
Sub a=4 back in equation:
(y+4)^2=4a(x+2)
-->(y+4)^2=(4*4)(x+2)
-->(y+4)^2=16(x+2)
Therefore you are correct :)
The general formula(s) for translated parabolas:
(y-k)^2=4a(x-h), with focus (h+a,k), and directrix at x=h-a.
OR
(x-h)^2=4a(y-k), focus (h,k+a) and directrix y=k-a
(so the vertex is (-2,-4), so h=-2 and k=-4.)
Since the directrix is "x" equals -6, it tells us that we should use the first formula for a translated parabola (shown above).
Sub (h,k) points into the equation
---> (y--4)^2=4a(x--2)
--->(y+4)^2=4a(x+2)
To find a, use the equation when finding a directrix (x=h-a)
x is given, (-6) and we know h.
-6=-2-a
a=4
Sub a=4 back in equation:
(y+4)^2=4a(x+2)
-->(y+4)^2=(4*4)(x+2)
-->(y+4)^2=16(x+2)
Therefore you are correct :)