   Chapter 10.1, Problem 14E

Chapter
Section
Textbook Problem

# Sketching a Parabola In Exercises 7–14, find the vertex, focus, and directrix of the parabola, and sketch its graph. y 2 + 4 y + 8 x − 12 = 0

To determine

To graph: The parabola y2+4y+8x12=0 and find its vertex, focus and directrix.

Explanation

Given:

The provided equation of parabola is:

y2+4y+8x12=0

Graph:

Consider the provided equation of parabola:

y2+4y+8x12=0

Convert the provided equation to standard form:

y2+4y+4+8x12=4(y+2)2=8(x2)

Now, the standard equation of a parabola with vertex (h,k) and directrix x=hp on a horizontal axis is given by:

(yk)2=4p(xh) …… (1)

And the coordinates of focus are:

(h+p,k)

Compare the equation (1) with the provided equation of parabola:

(y+2)2=8(x2)

Value of k is 2, value of p is 2 and value of h is 2.

Thus, substitute 2 for h, 2 for p and 2 for k, the focus of the provided parabola is (0,2).

And also, since h=2 and k=2, the vertex of the parabola is (2,2)

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