How do you write the parabola 2y^2+4y+x-8=02y2+4y+x8=0 in standard form and find the vertex, focus, and directrix?

2 Answers
Jun 1, 2018

You try to make a perfect square!

Explanation:

2(y^2 +2y +1 -1) +x -8 = 02(y2+2y+11)+x8=0

2(y+1)^2-2+x-8=02(y+1)22+x8=0

2(y+1)^2+x-10=02(y+1)2+x10=0

2(y+1)^2 = -(x-10)2(y+1)2=(x10)

(y+1)^2 = -1/2(x-10)(y+1)2=12(x10)

(y+1)^2 = 4(-1/8)(x-10)(y+1)2=4(18)(x10)

Jun 1, 2018

This is a parabola that opens to the left, therefore, its standard form is:

x = ay^2+by+c , a < 0x=ay2+by+c,a<0

The formulas for the requested items will be given in the explanation.

Explanation:

Given: 2y^2+4y+x-8=02y2+4y+x8=0

To write in standard form, we add -2y^2-4y+82y24y+8 to both sides:

x = -2y^2-4y+8x=2y24y+8

The above is standard form where a = -2a=2, b = -4b=4, and c = 8c=8

The vertex is a point (h,k)(h,k)

The formula for the y-coordinate of the vertex is:

k = -b/(2a)k=b2a

Substitute in the values, a = -2a=2 and b = -4b=4

k = -(-4)/(2(-2)k=42(2)

k = -1k=1

The formula for the x-coordinate of the vertex is:

h = ak^2+b(k) + ch=ak2+b(k)+c

Substitute in the values, h = -1h=1, a = -2a=2, b = -4b=4, and c = 8c=8

h = -2(-1)^2-4(-1) + 8h=2(1)24(1)+8

h =10h=10

The vertex is (10,-1)(10,1)

The focal distance ff is:

f = 1/(4a)f=14a

f = 1/(4(-2))f=14(2)

f = -1/8f=18

The formula for the focus is:

(h+f, k)(h+f,k)

Substitute in values h = 10h=10, f = -1/8f=18, and k = -1k=1

(10-1/8, 10)(1018,10)

The focus is (79/8, -1)(798,1)

The formula for the equation of the directrix is:

x = h-fx=hf

Substitute in values h = 10h=10 and f = -1/8f=18:

x = 10 - -1/8x=1018

x = 81/8x=818

The above is the equation of the directrix.

The following is a drawing of the parabola, the vertex, the focus, and the directrix:

![www.desmos.com/calculator](useruploads.socratic.orguseruploads.socratic.org)