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A parabola can be drawn given a focus of (8,10) and a directrix of y=4. What can be said about the parabola?Asked by idk
A parabola can be drawn given a focus of (5,−10) and a directrix of y=6. What can be said about the parabola?
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Answered by
oobleck
we have p = -8, so
(x-5)^2 = -32(y+2)
(x-5)^2 = -32(y+2)
Answered by
Bot
- The vertex of the parabola is halfway between the focus and the directrix, which is at (5, -2).
- The parabola will open downwards because the directrix is above the focus.
- The distance between the vertex and the focus (or the directrix) is called the "p" value of the parabola. In this case, the distance between the vertex and the directrix is 8 units, so the p value is -8 (since the parabola opens downwards). Therefore, the equation of the parabola can be written as: (y + 2) = -1/8(x - 5)^2.
- The parabola will open downwards because the directrix is above the focus.
- The distance between the vertex and the focus (or the directrix) is called the "p" value of the parabola. In this case, the distance between the vertex and the directrix is 8 units, so the p value is -8 (since the parabola opens downwards). Therefore, the equation of the parabola can be written as: (y + 2) = -1/8(x - 5)^2.
Answered by
Bot
I apologize, as an AI language model, I made a mistake in my previous response. You are correct, since the parabola opens downwards, the value of p is negative. Therefore, the equation of the parabola can be written as:
(x - 5)^2 = -32(y + 2)
This equation is in vertex form, with vertex at (5, -2) and p = -8.
(x - 5)^2 = -32(y + 2)
This equation is in vertex form, with vertex at (5, -2) and p = -8.
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