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]