Question 1)
The given parabola is:
[tex]y= \frac{1}{8} x^{2} +4x+20 [/tex]
First we need to convert this equation to the general form similar to:
[tex]4p(y-k)=(x-h)^{2} [/tex]
So, the given equation will be:
[tex]y= \frac{1}{8} x^{2} +4x+20 \\ \\
y= \frac{1}{8}( x^{2} +32x)+20 \\ \\
y= \frac{1}{8}( x^{2} +32x+256)+20- \frac{1}{8}(256) \\ \\
y = \frac{1}{8}(x+16)^{2}-12 \\ \\
y+12= \frac{1}{8}(x+16)^{2} \\ \\
8(y+12)=(x+16)^{2} \\ \\
4*2(y-(-12))=(x-(-16))^{2} [/tex]
Comparing given equation with the general equation, we can write:
p = 2
h = -12
k= -16
The focus of the parabola with squared x terms lies at (h , k+p)
Using the values, we get the focus point of the given parabola (-16, -10)
Question 2.
The equation of the parabola is:
[tex](y-1)^{2}=16(x+3) \\ \\
(y-1)^{2}=4*4(x+3)[/tex]
This means:
p=4
h = -3
k =1
The directrix of the parabola with squared y term has the equation of the form:
x = h - p
Using the values, we get
x= - 3 - 4 = -7
So, the equation of the directrix of the parabola will be x = -7
