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Find the equation of the parabola with its focus at (6,2) and its directrix y=0.

a) y=(x−6) ^2 +2
b) y=−(x−6) ^2 +2
c) y=(x−6)+2
d) y=−(x−6)+2

User Bala
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1 Answer

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Final answer:

The equation of the parabola with its focus at (6,2) and its directrix y=0 is (x - 6)² = 8(y - 2).

Step-by-step explanation:

The equation of the parabola with its focus at (6,2) and its directrix y=0 can be found using the standard form of the equation of a parabola. The standard form is given by:

(x - h)² = 4p(y - k)

In this case, the vertex (h,k) is (6,2) and the distance from the vertex to the focus is given by p. Since the directrix is y=0, the distance from the vertex to the directrix is also given by p. We can substitute the values into the standard form to get the equation of the parabola:

(x - 6)² = 4p(y - 2)

Since p is the same for both the focus and the directrix, we can solve for p by finding the distance from the vertex to the focus or the distance from the vertex to the directrix. In this case, the distance from the vertex (6,2) to the directrix y=0 is 2, so p = 2. Substituting p = 2 into the equation gives us:

(x - 6)² = 8(y - 2)

So the equation of the parabola with its focus at (6,2) and its directrix y=0 is (x - 6)² = 8(y - 2).

User Dzmitry Lazerka
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