Since the axis is horizontal, the general equation is
x = a(y-k)^2 + h
Since the vertex is at y=k, k = h and that means
x = a(y-k)^2 + k
Using the two points, we have
a(-2-k)^2 + k = 6
a(4-k)^2 + k = 3
solve those, and you wind up with
x = 1/4 (y-2)^2 + 2
x = 5/36 (y-14/5) + 14/5
so, there are two parabolas which meet the conditions.
See the graphs at
http://www.wolframalpha.com/input/?i=plot+x+%3D+1%2F4+(y-2)%5E2+%2B+2,+x+%3D+5%2F36+(y-14%2F5)%5E2+%2B+14%2F5,+y%3Dx
Plss... help me.. what is the equation of parabola with vertex on the line y=x, axis parallel to Ox, and passing through (6,-2) and (3,4)? thanks!
2 answers
vertex at point (v,v)
axis parallel to x, opening right
(y-v)^2 = 4a(x-v)
so
(-2-v)^2 = 4 a (6-v)
and
( 4-v)^2 = 4 a (3-v)
======================now plug and chug
4 + 4v + v^2 = 24 a - 4 a v
16- 8v + v^2 = 12 a - 4 a v
-----------------------------
-12 +12v = 12 a or a = v-1
4 + 4 v + v^2=24(v-1)-4v(v-1)
v^2+4v+4 = 24v -24 -4v^2 +4 v
5v^2 -24 v +28 = 0
(5v-14)(v-2)=0
v = 2 or v = 14/5
if v = 2
a = 1
(y-2)^2= 4(1)(x-2)
y^2 -4 y + 4 = 4 x -8
4x = y^2 -4y + 12 check (it works)
I will let you try v = 14/5 :)
axis parallel to x, opening right
(y-v)^2 = 4a(x-v)
so
(-2-v)^2 = 4 a (6-v)
and
( 4-v)^2 = 4 a (3-v)
======================now plug and chug
4 + 4v + v^2 = 24 a - 4 a v
16- 8v + v^2 = 12 a - 4 a v
-----------------------------
-12 +12v = 12 a or a = v-1
4 + 4 v + v^2=24(v-1)-4v(v-1)
v^2+4v+4 = 24v -24 -4v^2 +4 v
5v^2 -24 v +28 = 0
(5v-14)(v-2)=0
v = 2 or v = 14/5
if v = 2
a = 1
(y-2)^2= 4(1)(x-2)
y^2 -4 y + 4 = 4 x -8
4x = y^2 -4y + 12 check (it works)
I will let you try v = 14/5 :)
1 answer