MATH SOLVE

4 months ago

Q:
# Which equation represents a parabola with a focus at (0,-2) and a directrix of y=6?

Accepted Solution

A:

The focus is:

[tex]F(0,-2)[/tex]

Given that the directrix is parallel to the x-axis, then the ordinary equation is given by:

[tex](x-h)^{2} = 4p(y-k)[/tex]

We need to find V(h,k), being V the vertex.

We know that these distances are always the same, namely:

│FV│ = │VD│

Being D the directrix. Given that the focus F is on the y-axis and the directrix is parallel to the x-axis, then the vertex V will also be on this axis, so h = 0.

As │FV│ = │VD│, then:

[tex]k = \frac{6-2}{2}[/tex], that is the middle point of the segment FD, so:

V(0,2)

Now │FV│= │p│= │2-(-2)│=4

Given that the vertex and focus are below the directrix, then the parabola open down, therefore: [tex]p\ \textless \ 0[/tex]

Lastly, the equation is:

[tex]x^{2} = -4(4)(y-2) = -16y+32[/tex]

[tex]y = -\frac{ x^{2} }{16} + 2[/tex]

[tex]F(0,-2)[/tex]

Given that the directrix is parallel to the x-axis, then the ordinary equation is given by:

[tex](x-h)^{2} = 4p(y-k)[/tex]

We need to find V(h,k), being V the vertex.

We know that these distances are always the same, namely:

│FV│ = │VD│

Being D the directrix. Given that the focus F is on the y-axis and the directrix is parallel to the x-axis, then the vertex V will also be on this axis, so h = 0.

As │FV│ = │VD│, then:

[tex]k = \frac{6-2}{2}[/tex], that is the middle point of the segment FD, so:

V(0,2)

Now │FV│= │p│= │2-(-2)│=4

Given that the vertex and focus are below the directrix, then the parabola open down, therefore: [tex]p\ \textless \ 0[/tex]

Lastly, the equation is:

[tex]x^{2} = -4(4)(y-2) = -16y+32[/tex]

[tex]y = -\frac{ x^{2} }{16} + 2[/tex]