Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar.The focus of a parabola is (0,–1). The directrix is the line y = 0 What is the equation of the parabola in vertex form?- k²thIn the equation =the value of pisThe vertex of the parabola is the pointThe equation of this parabola in vertex form is y=12 - 1

Type the correct answer in each box Use numerals instead of words If necessary use for the fraction barThe focus of a parabola is 01 The directrix is the line y class=

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The Solution:

Given the equation of the parabola in vertex form:

[tex]y=ax^2-1[/tex]

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

[tex]\begin{gathered} \text{ Focus: (h,K+p)=(0,-1)} \\ h=0 \\ k+p=-1\ldots\text{eqn}(1) \end{gathered}[/tex]

By directrix:

[tex]\begin{gathered} y=k-p \\ 0=k-p \\ k-p=0\ldots eqn(2) \end{gathered}[/tex]

Solving eqn(1) and eqn(2) simultaneously, we get

[tex]\begin{gathered} 2k=-1 \\ \\ k=-\frac{1}{2}=-0.5 \end{gathered}[/tex]

So, the directrix is:

[tex](h,k)=(0,-0.5)[/tex]

So, the equation of the parabola is

[tex]\begin{gathered} y=\frac{1}{4(-0.5)}x^2-0.5 \\ \\ y=-0.5x^2-0.5 \end{gathered}[/tex]

So, the value of p is -0.5

Thus, the equation of the parabola in vertex form is:

[tex]y=-0.5x^2-0.5[/tex]

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