Shazer2
Member
- Joined
- Feb 16, 2012
- Messages
- 439
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- Male
- HSC
- 2013
A parabola has a focus at (0, 4) and its vertex is at (0, 2). Find the equation of the parabola.
I graphed the focus and vertex, and since the vertex is the midpoint between the focus and directrix we can say the directrix is at y=0. So we can say the focal length is 2, a=2. Then we can sub that it in for a regular parabola like so, x^2=4(2)(y) and that results in x^2 = 8y. However the answer is x^2-8y+16=0. I'm not sure how they got that. Any help?
I graphed the focus and vertex, and since the vertex is the midpoint between the focus and directrix we can say the directrix is at y=0. So we can say the focal length is 2, a=2. Then we can sub that it in for a regular parabola like so, x^2=4(2)(y) and that results in x^2 = 8y. However the answer is x^2-8y+16=0. I'm not sure how they got that. Any help?