Question

Write the equation in vertex form and then find the vertex, focus, and directrix of a parabola with equation x=y^2+18y-2

Answer 1

Answer:

Part 1) The vertex is the point (-83,-9)

Part 2) The focus is the point (-82.75,-9)

Part 3) The directrix is $x=-83.25$

Step-by-step explanation:

step 1

Find the vertex

we know that

The equation of a horizontal parabola in the standard form is equal to

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

where

p≠ 0.

(h,k) is the vertex

(h + p, k) is the focus

x=h-p is the directrix

In this problem we have

$x=y^{2} +18y-2$

Convert to standard form

$x+2=y^{2} +18y$

$x+2+81=y^{2} +18y+81$

$x+83=(y+9)^{2}$

so

This is a horizontal parabola open to the right

(h,k) is the point (-83,-9)

so

The vertex is the point (-83,-9)

step 2

we have

$x+83=(y+9)^{2}$

Find the value of p

$4p=1$

$p=1/4$

Find the focus

(h + p, k) is the focus

substitute

(-83+1/4,-9)

The focus is the point (-82.75,-9)

step 3

Find the directrix

The directrix of a horizontal parabola is

$x=h-p$

substitute

$x=-83-1/4$

$x=-83.25$