(2x+1)^2-2(2x^2-1) Simplify
1 answer:
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Answer:
y = -1/10x^2 +2.5
Step-by-step explanation:
The distance from focus to directrix is twice the distance from focus to vertex. The focus-directrix distance is the difference in y-values:
-1 -4 = -5
So, the distance from focus to vertex is p = -5/2 = -2.5. This places the focus 2.5 units below the vertex. Then the vertex is at (h, k) = (0, -1) +(0, 2.5) = (0, 1.5).
The scale factor of the parabola is 1/(4p) = 1/(4(-2.5)) = -1/10. Then the equation of the parabola is ...
y = (1/(4p))(x -h) +k
y = -1/10x^2 +2.5
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You can check the graph by making sure the focus and directrix are the same distance from the parabola everywhere. Of course, if the vertex is halfway between focus and directrix, the distances are the same there. Another point that is usually easy to check is the point on the parabola that is even with the focus. It should be as far from the focus as it is from the directrix. In this parabola, the focus is 5 units from the directrix, and we see the points on the parabola at y=-1 are 5 units from the focus.
