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[tex]\sqrt{(x_{0} - 4)^{2} + (y_{0} - (-7))^{2}} = |y_{0} - (-15)|[/tex]
[tex](\sqrt{(x_{0} - 4)^{2} + (y_{0} + 7)^{2})^{2}} = (y_{0} + 15)^{2}[/tex]
[tex](x_{0} - 4)^{2} + (y_{0} + 7)^{2} = (y_{0} + 15)^{2}[/tex]
[tex](x_{0}^{2} - 8x_{0} + 16) + (y_{0}^{2} + 14y_{0} + 49) = y_{0}^{2} + 30y_{0} + 225[/tex]
[tex]x_{0}^{2} + y_{0}^{2} - 8x_{0} + 14y_{0} + 16 + 49 = y_{0}^{2} + 30y_{0} + 225[/tex]
[tex]x_{0}^{2} + y_{0}^{2} - 8x_{0} + 14y_{0} + 65 = y_{0}^{2} + 30y_{0} + 225[/tex]
[tex]x_{0}^{2} - 8x_{0} - 16y_{0} - 160 = 0[/tex]
[tex]16y_{0} = x_{0}^{2} - 8x_{0} - 160[/tex]
[tex]\frac{16y_{0}}{16} = \frac{x_{0}^{2} - 8x_{0} - 160}{16}[/tex]
[tex]y_{0} = \frac{1}{16}x_{0}^{2} - \frac{1}{2}x_{0} - 10[/tex]
[tex]y = \frac{1}{16}x^{2} - \frac{1}{2}x - 10[/tex]

Answer:

The equation of the parabola with a focus at (4,-7) and a directrix of y=-15 is [tex]y=\frac{x^{2}}{16}-\frac{x}{2}-10[/tex]

Step-by-step explanation:

we need to drive the equation of the parabola with a focus at ( 4, -7) and a directrix of y= -15

From the given focus ( 4, -7) and equation of directrix  y = - 15 calculate p

[tex]p=\frac{1}{2}(y_0-y)[/tex]

where [tex]y_0[/tex] is is ordinate of focus and y is equation of directrix.

[tex]p=\frac{1}{2}(-7-(-15))[/tex]

[tex]p=\frac{1}{2}(-7+15)[/tex]

[tex]p=\frac{1}{2}(8)[/tex]

[tex]p=4[/tex]

Calculate the vertex (h,k)

[tex]h=4\; \text{and}\; k=\frac{-7+(-15)}{2}=-11[/tex]

vertex (h,k) =(4,-11)

Since, vertex  form is :

[tex](x-h)^{2}=4p(y-k)[/tex]       (positive 4p shows it open upward)

[tex](x-4)^{2}=4(4)(y-(-11))[/tex]

[tex](x-4)^{2}=16(y+11)[/tex]

[tex]x^{2}+16-8x=16y+176[/tex]

subtract both the sides by 176,

[tex]x^{2}-8x-160=16y[/tex]

Divide both the sides in above by 16,

[tex]\frac{x^{2}}{16}-\frac{8x}{16}-\frac{160}{16}=y[/tex]

[tex]\frac{x^{2}}{16}-\frac{x}{2}-10=y[/tex]

Hence, the equation of the parabola with a focus at (4,-7) and a directrix of y=-15 is  [tex]y=\frac{x^{2}}{16}-\frac{x}{2}-10[/tex]

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