What is the focus and vertex of the parabola described by #x^2+4x+4y+16=0 #?

Question

Genesis Allen · Accepted Answer

#"focus "=(-2,-4)," vertex "=(-2,-3)# #"the equation of a vertically opening parabola is"# #•color(white)(x)(x-h)^2=4a(y-k)# #"where "(h,k)" are the coordinates of the vertex and a"# #"is the distance from the vertex to the focus/directrix"# #• " if "4a>0" then opens upwards"# #• " if "4a<0" then opens downwards"# #"rearrange "x^2+4x+4y+16=0" into this form"# #"using the method of "color(blue)"completing the square"# #x^2+4xcolor(red)(+4)=-4y-16color(red)(+4)# #(x+2)^2=-4(y+3)# #color(magenta)"vertex "=(-2,-3)# #4a=-4rArra=-1# #color(purple)" focus "=(-2,-3-1)=(-2,-4)# graph{x^2+4x+4y+16=0 [-10, 10, -5, 5]}

Eli Assouline · Answer

undefined To find the focus and vertex of the parabola described by $x^2 + 4x + 4y + 16 = 0$, first, rewrite the equation in the standard form $y = a(x - h)^2 + k$. Then, use the formula for the focus of a parabola, which is $(h, k + \frac{1}{4a})$, and the vertex formula, which is $(h, k)$.

Given the equation $x^2 + 4x + 4y + 16 = 0$, rearrange it to isolate $y$ to get $y = -\frac{1}{4}(x^2 + 4x + 16)$. Then, complete the square for the $x$ terms:

$$x^2 + 4x + 16 = (x + 2)^2$$

So, $y = -\frac{1}{4}(x + 2)^2$.

Comparing this to the standard form $y = a(x - h)^2 + k$, you can see that $a = -\frac{1}{4}$, $h = -2$, and $k = 0$.

Now, the vertex is $(h, k) = (-2, 0)$.

The focus is $(h, k + \frac{1}{4a}) = (-2, 0 + \frac{1}{4*(-\frac{1}{4})}) = (-2, 1)$.

So, the vertex of the parabola is (-2, 0) and the focus is (-2, 1).