Question

1. The equation of a parabola is given.

Answer 1

Question 1)

The given parabola is:

$y= \frac{1}{8} x^{2} +4x+20$

First we need to convert this equation to the general form similar to:

$4p(y-k)=(x-h)^{2}$

So, the given equation will be:

$y= \frac{1}{8} x^{2} +4x+20 \\ \\ y= \frac{1}{8}( x^{2} +32x)+20 \\ \\ y= \frac{1}{8}( x^{2} +32x+256)+20- \frac{1}{8}(256) \\ \\ y = \frac{1}{8}(x+16)^{2}-12 \\ \\ y+12= \frac{1}{8}(x+16)^{2} \\ \\ 8(y+12)=(x+16)^{2} \\ \\ 4*2(y-(-12))=(x-(-16))^{2}$

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:
$(y-1)^{2}=16(x+3) \\ \\ (y-1)^{2}=4*4(x+3)$

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