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Write an equation for a parabola with a focus of (-4,0) and a directrix of y=4.a. y=1/4(x-4)^2+2b. y=-1/8(x+4)^2+2c. y=1/8(x+4)^2+2d. y=-1/4(x+4)^2+2

Respuesta :

[tex]\begin{gathered} The_{\text{ }}directrix_{\text{ }}is: \\ y=k-p \\ The_{\text{ }}focus_{\text{ }}is: \\ F=(h,k\pm p) \end{gathered}[/tex]

So, using the information provided by the problem:

[tex]\begin{gathered} 4=k-p_{\text{ }}(1) \\ 0=k+p_{\text{ }}(2) \\ (1)+(2) \\ 4=2k \\ k=2 \\ p=-2 \end{gathered}[/tex]

Therefore:

[tex]\begin{gathered} 4p(y-k)=(x-h)^2 \\ y=\frac{1}{4p}(x-h)^2+k \\ y=\frac{1}{4(-2)}(x-(-4))^2+2 \\ y=-\frac{1}{8}(x+4)^2+2 \end{gathered}[/tex]

Answer:

b. y=-1/8(x+4)^2+2

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