Verified Answer :

Panoyin

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

Answer:

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

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

\(p=\frac{1}{2}(y_0-y)\)

where \(y_0\) is is ordinate of focus and y is equation of directrix.

\(p=\frac{1}{2}(-7-(-15))\)

\(p=\frac{1}{2}(-7+15)\)

\(p=\frac{1}{2}(8)\)

\(p=4\)

Calculate the vertex (h,k)

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

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

Since, vertex  form is :

\((x-h)^{2}=4p(y-k)\)       (positive 4p shows it open upward)

\((x-4)^{2}=4(4)(y-(-11))\)

\((x-4)^{2}=16(y+11)\)

\(x^{2}+16-8x=16y+176\)

subtract both the sides by 176,

\(x^{2}-8x-160=16y\)

Divide both the sides in above by 16,

\(\frac{x^{2}}{16}-\frac{8x}{16}-\frac{160}{16}=y\)

\(\frac{x^{2}}{16}-\frac{x}{2}-10=y\)

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

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