A vertical parabola with vertex (1,-4) and focus (1,-2) is given. It is required to find the standard form of the equation for the parabola.
Recall that the equation of a Vertical Parabola with vertex (h,k) and focus (h,k+1/4a) is given as:
[tex]y=a(x-h)^2+k[/tex]Compare the given vertex (1,-4) with the form (h,k). It follows that:
[tex]h=1,k=-4[/tex]Compare the given focus (1,-2) with the form (h,k+1/4a). It follows that:
[tex]k+\frac{1}{4a}=-2[/tex]Substitute k=-4 into the equation and find the value of a:
[tex]\begin{gathered} -4+\frac{1}{4a}=-2 \\ \Rightarrow\frac{1}{4a}=-2+4 \\ \Rightarrow\frac{1}{4a}=2 \\ \Rightarrow4a=\frac{1}{2} \\ \Rightarrow a=\frac{1}{8} \end{gathered}[/tex]Substitute k=-4, h=1, and a=1/8 into the equation of a vertical parabola:
[tex]y=\frac{1}{8}(x-1)^2+(-4)[/tex]Rewrite the equation in standard form as follows:
[tex]\begin{gathered} y+4=\frac{1}{8}(x-1)^2 \\ \Rightarrow8(y+4)=(x-1)^2 \\ \text{ Swap the sides of the equation:} \\ \Rightarrow(x-1)^2=8(y+4) \end{gathered}[/tex]Recall that the Directrix of a vertical parabola is represented by the equation:
[tex]y=k-\frac{1}{4a}[/tex]Substitute k=-4 and a=1/8 into the equation of directrix:
[tex]\begin{gathered} y=-4-\frac{1}{4(\frac{1}{8})}=-4-\frac{1}{\frac{1}{2}}=-4-2=-6 \\ \Rightarrow y=-6 \end{gathered}[/tex]Hence, the correct answer is (x-1)²=8(y+4); y=-6.
The correct answer is (x-1)² = 8(y+4) ; y = - 6. (second option)
