Derive the equation of the parabola with a focus at (0, −4) and a directrix of y = 4.


A.) f(x) = −16x^2

B.) f(x) = 16x^2

C.) f(x) = −1/16 x^2

D.) f(x) = 1/16 x^2

Verified Answer :

\(\sqrt{(x_{0} - 0)^{2} + (y_{0} - (-4))^{2}} = |y_{0} - 4|\)
\(\sqrt{(x_{0} - 0)^{2} + (y_{0} + 4)^{2}} = |y_{0} - 4|\)
\((x_{0} - 0)^{2} + (y_{0} + 4)^{2} = (y_{0} - 4)^{2}\)
\((x_{0})^{2} + (y_{0} + 4)^{2} = (y_{0} - 4)^{2}\)
\(x_{0}^{2} + (y_{0}^{2} + 4y_{0} + 4y_{0} + 16) = y_{0}^{2} - 4y_{0} - 4y_{0} + 16\)
\(x_{0}^{2} + (y_{0}^{2} + 8y_{0} + 16) = y_{0}^{2} - 8y_{0} + 16\)
\(x_{0}^{2} + y_{0}^{2} + 8y_{0} + 16 = y_{0}^{2} - 8y_{0} + 16\)
\(x_{0}^{2} + 16y_{0} = 0\)
\(16y_{0} = -x_{0}^{2}\)
\(\frac{16y_{0}}{16} = \frac{-x_{0}^{2}}{16}\)
\(y_{0} = -\frac{1}{16}x_{0}^{2}\)
\(y = -\frac{1}{16}x^{2}\)

The answer is C.