   Chapter 10.5, Problem 7E

Chapter
Section
Textbook Problem

Find the vertex, focus, and directrix of the parabola and sketch its graph.7. y2 + 6y + 2x + 1 = 0

To determine

To Find: The vertex, focus, and directrix of the parabola for the equation y2+6y+2x+1=0.

Explanation

Given:

The parabola equation is as follows.

y2+6y+2x+1=0

Rewrite the equation as,

y2+6y+2x+1=0y2+6y+2x+(98)=0y2+6y+9=2x+8

Let Factories the above equation.

(y+3)2=2x+8

Rearrange the equation again.

We get,

(y+3)2=2x+812(y+3)2=12(2x+8)12(y+3)2=(x+4)

Calculation:

Compute the vertex of the parabola.

The equation of the vertex is,

(yk)2=4p(xh)12(y+3)2=(x4)

Then, the vertex is said to be (h,k)

Therefore, the vertex is (4,3)

Compute the focus of the parabola.

Compare the parabola equation with the below equation,

y2=4px

Let, take the parabola equation as,

12(y+3)2=(x+4)14p=121=2pp=12

Compute the focus of the parabola.

Focus=(h+p,k)

Substitute the 4 for h, 3 for k and 12 for  p

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