7. Focus: $$ \left(0,-\frac{3}{2}\right) $$

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UpStudy AI · Accepted Answer

We assume the parabola has a vertical axis of symmetry with its vertex at the origin. In the standard form for a vertical parabola with vertex $$(0,0)$$ the equation is  
$$
x^2 = 4py,
$$  
where $$p$$ is the distance from the vertex to the focus.

Since the focus is given as  
$$
\left(0,-\frac{3}{2}\right),
$$  
it follows that  
$$
p = -\frac{3}{2}.
$$

Substituting $$p = -\frac{3}{2}$$ into the standard equation gives  
$$
x^2 = 4\left(-\frac{3}{2}\right)y.
$$

Simplify the multiplication:  
$$
4\left(-\frac{3}{2}\right) = -6.
$$

Thus the equation becomes  
$$
x^2 = -6y.
$$

For completeness, note that the directrix of the parabola is a horizontal line given by  
$$
y = -p.
$$

Since $$p = -\frac{3}{2}$$, we have  
$$
y = -\left(-\frac{3}{2}\right) = \frac{3}{2}.
$$

In summary, the parabola with focus $$\left(0,-\frac{3}{2}\right)$$ (and vertex at the origin) has:

- Equation:  
  $$
  x^2 = -6y,
  $$
- Focus:  
  $$
  \left(0,-\frac{3}{2}\right),
  $$
- Directrix:  
  $$
  y = \frac{3}{2}.
  $$
The parabola with focus at $$\left(0,-\frac{3}{2}\right)$$ and vertex at the origin has the equation $$x^2 = -6y$$.

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