- Focus: $F(-3,-1)$
- Axis of symmetry of the form $y=p$, so directrix of the form $x=p$
- Because the parabolas open to the left, $p>-3$
My computations are as follows:
$(x+3)^2 + (y+1)^2=(x-p)^2$ with $p>-3$
$x^2+6x+9 + (y+1)^2= x^2-2px+p^2$
$(y+1)^2=-2px+p^2-6x-9$
$(y+1)^2=-2x(p+3)+(p+3)(p-3)$
$(y+1)^2=-2(p+3)(x-\frac{p-3}{2})$
Book answer:
$(y+1)^2=-4(h+3)(x-h)$ with $h>-3$
Why is my answer different? Both yield the same parabola when $p=5$ and $h=1$, or when $p=1$ and $h=-1$
What approach could yield the book answer?