Answer:
![\displaystyle y=-\frac{1}{10}x^2-\frac{4}{5}x-\frac{11}{10}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D-%5Cfrac%7B1%7D%7B10%7Dx%5E2-%5Cfrac%7B4%7D%7B5%7Dx-%5Cfrac%7B11%7D%7B10%7D)
Step-by-step explanation:
By definition, any point (<em>x, y</em>) on the parabola is equidistant from the focus and the directrix.
The distance between a point (<em>x, y</em>) on the parabola and the focus can be described using the distance formula:
![d=\sqrt{(x-(-4))^2+(y-(-2))^2](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x-%28-4%29%29%5E2%2B%28y-%28-2%29%29%5E2)
Simplify:
![d=\sqrt{(x+4)^2+(y+2)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x%2B4%29%5E2%2B%28y%2B2%29%5E2%7D)
Since the directrix is an equation of <em>y</em>, we will use the <em>y-</em>coordinate. The vertical distance between a point (<em>x, y</em>) on the parabola and the directrix can be described using absolute value:
![d=|y-3|\text{ or } |3-y|](https://tex.z-dn.net/?f=d%3D%7Cy-3%7C%5Ctext%7B%20or%20%7D%20%7C3-y%7C)
The two equations are equivalent. Therefore:
![\sqrt{(x+4)^2+(y+2)^2}=|y-3|](https://tex.z-dn.net/?f=%5Csqrt%7B%28x%2B4%29%5E2%2B%28y%2B2%29%5E2%7D%3D%7Cy-3%7C)
Solve for <em>y</em>. We can square both sides. Since anything squared is positive, we can remove the absolute value:
![(x+4)^2+(y+2)^2 = (y-3)^2](https://tex.z-dn.net/?f=%28x%2B4%29%5E2%2B%28y%2B2%29%5E2%20%3D%20%28y-3%29%5E2)
Expand:
![(x^2+8x+16)+(y^2+4y+4)=(y^2-6y+9)](https://tex.z-dn.net/?f=%28x%5E2%2B8x%2B16%29%2B%28y%5E2%2B4y%2B4%29%3D%28y%5E2-6y%2B9%29)
Isolate:
![x^2+8x+11=-10y](https://tex.z-dn.net/?f=x%5E2%2B8x%2B11%3D-10y)
Divide both sides by -10. Hence, our equation is:
![\displaystyle y=-\frac{1}{10}x^2-\frac{4}{5}x-\frac{11}{10}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D-%5Cfrac%7B1%7D%7B10%7Dx%5E2-%5Cfrac%7B4%7D%7B5%7Dx-%5Cfrac%7B11%7D%7B10%7D)