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soldier1979 [14.2K]
3 years ago
11

A parabola can be drawn given a focus of (3, 1) and a directrix of x = -5. What can

Mathematics
1 answer:
hjlf3 years ago
7 0

Answer:

The statements that can be made about the parabola are;

The parabola opens towards the right

The equation of the parabola is (y - 1)² = 16 × (x + 1)

The parabola has a vertex at (-1, 1)

Step-by-step explanation:

The given focus of the parabola is (3, 1)

The directrix of the parabola, x = -5

Therefore, from the location of the directrix (on the x-axis) and the location of the focus relative to the directrix towards the positive direction of the x-axis relative to the directrix, the parabola opens towards the left of the coordinate chart

The general equation of the focus = (h + p, k)

The general equation of the directrix, x = h - p

Comparing with the values of the focus and the directrix of the given parabola, we have;

(3, 1) = (h + p, k)

-5 = h - p

Therefore, we get;

k = 1

h + p = 3...(1)

h - p = -5...(2)

Adding equation (1) to equation (2) gives;

h + h + p - p = 3 + (-5)

2·h = -2

h = -2/2 = -1

h = -1

From equation (1), we get;

h + p = -1 + p = 3

∴ p = 3 + 1 = 4

p = 4

The vertex of the parabola, (h, k) = (-1, 1)

The equation of the parabola in the form (y - k)² = 4·p·(x - h) is therefore, presented as follows;

The equation of the parabola = (y - 1)² = 4 × 4 × (x - (-1))

(y - 1)² = 16 × (x - (-1))

The equation of the parabola is (y - 1)² = 16 × (x + 1).

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