Write the equation for a parabola with a vertex at (-1,3) and the equation of the directrix y=1

1. f(-1,3) directrix=y=3 Also, find the vertex. 2. Draw what you know on a graph: y=1 is a line parallel to the X-axis and going through the point (-1,3). The focus is at (-1,3) above the directrix. The vertex should be above the focus point and half way the distance from the directrix line and the focus point. Using y values you can say the total distance is 3-1=2 units Then half of that would be one unit. Therefore, the vertex is at (0, 2), which is in the middle distance of the directrix and the focus point. 3. The parabola is going to open upwards because the parabola will open away from the directrix. Also, the p=distance from vertex to focus is +1 unit upward, so p= +1. 4. Use this standard equation: (X-h)2 = 4p(y-k). V(h,k)=(0,2) h=0 and k=2. 5. (X-0)2 = 4(1)(y-2) X2 =4(y-2) Answer

astha8579 astha8579 · Answer 2 · 2022-01-08T12:58:23+01:00

The equation of the parabola with a vertex at (-1, 3) and the equation of the directrix y = 1 is: [tex](x+1)^2 = 8(y-3)\\[/tex]

Given that:

Vertex of the parabola: (-1,3)

Equation of the directrix: y = 1

The standard form [tex](x-h)^2 = 4p(y-k)[/tex] is the equation of a parabola with (h,k) as its vertex's coordinates and y = k - p as its equation of directrix.

Since the given vertex is on (-1,3), thus h = -1 and k = 3

Then equation of directrix is:

[tex]y = k - p = 1\\k - p = 1\\3-p = 1\\p = 2[/tex]

Thus equation of the parabola is given as

[tex](x - (-1))^2 = 4 \times 2 \times (y - 3)\\(x+1)^2 = 8(y-3)\\[/tex]

The plot of that parabola is shown below:

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