What are the coordinates of the focus of the parabola shown below?y² +16y+ 4x + 4 =0

Answer :

Answer:

(14, -8)

Explanations:

The standard equation of a parabola is expressed as;

[tex]y^2=4ax[/tex]

Given the equation of a parabola y² +16y+ 4x + 4 =0

Using the completing the square method

[tex]\begin{gathered} (y^2+16y+(\frac{16}{2})^2)=-4x-4+(\frac{16}{2})^2 \\ (y^2+16y+8^2)=-4x-4+8^2 \\ (y+8)^2=-4x-4+64 \\ (y+8)^2=-4x+60 \\ (y+8)^2=-4(x-15) \\ (y-(-8))^2=4(-1)(x-15) \end{gathered}[/tex]

From the result, you can see that the vertex (h, k) is (15, -8) and a = -1

Determine the focus of the parabola

[tex]\begin{gathered} focus=(15+a,-8) \\ focus=(15+(-1),-8) \\ focus=(14,-8) \end{gathered}[/tex]

Hence the focus of the parabola is (14, -8)