Question
6. Find the equation of each parabola described below.
a) parabola with vertex (0,0) and the focus (0,7)
b) parabola with focus (-3,0) and directrix x=3
c) parabola with vertex (3,3) and directrix x=-1
d) parabola with focus (-2,-1) and directrix y=5
e) ysquared=-6x translated according to ((x,y) -->(x-2,y+4))
could someone help me with these??
a) parabola with vertex (0,0) and the focus (0,7)
b) parabola with focus (-3,0) and directrix x=3
c) parabola with vertex (3,3) and directrix x=-1
d) parabola with focus (-2,-1) and directrix y=5
e) ysquared=-6x translated according to ((x,y) -->(x-2,y+4))
could someone help me with these??
Answers
You need to know the following:
vertex at (h,k)
if it opens up or down:
4 a (y-k) = (x-h)^2
a negative a opens down
if it opens right or left:
4 a (x-h) = (y-k)^2
a negative a opens left
distance vertex to focus = a
distance vertex to directrix = a
ALWAYS SKETCH THEM
================================
for example a is vertex (0,0) so h = k = 0 and focus (0,7)
the sketch shows it with axis along the y axis opening up and distance a = 7
so
4 * 7 * y = x^2
or y = x^2/28
vertex at (h,k)
if it opens up or down:
4 a (y-k) = (x-h)^2
a negative a opens down
if it opens right or left:
4 a (x-h) = (y-k)^2
a negative a opens left
distance vertex to focus = a
distance vertex to directrix = a
ALWAYS SKETCH THEM
================================
for example a is vertex (0,0) so h = k = 0 and focus (0,7)
the sketch shows it with axis along the y axis opening up and distance a = 7
so
4 * 7 * y = x^2
or y = x^2/28
Note that the vertex is halfway between the focus and the directrix by the way.
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