   Chapter 10.5, Problem 16E

Chapter
Section
Textbook Problem

# Find the vertices and foci of the ellipse and sketch its graph.16. x2 + 3y2 + 2x − 12y + 10 = 0

To determine

To Find: The vertices and foci of the ellipse for the equation x2+3y2+2x12y+10=0 .

Explanation

The ellipse equation is as follows.

x2+3y2+2x12y+10=0 (1)

Let rewrite the equation as follows,

x2+2x+3y212y+10=0(x2+2x)+(3y212y)+10=0(x2+2x)+3(y24y)+10=0(x2+2x+11)+3(y24y+44)+10=0(x2+2x+1)1+3(y24y+4)12+10=0

(x2+2x+1)1+3(y24y+4)2=0(x+1)2+3(y2)2=3

Divide the equation by the value 3 ,

(x+1)23+3(y2)23=33(x+1)23+(y2)21=1

Then, compare the equation (1) with the standard equation of ellipse,

a>b

So, the foci of an ellipse are located on the y axis using equation below,

x2a2+y2b2=1

Calculation:

Compute the center of the ellipse using the equation.

(xh)2a2+(yk)2b2=1(x+1)23+(y2)21=1

Therefore, the center of the ellipse (h,k) is (1,+2) .

Substitute the value 3 for a2 and 1 for b2 .

a2=3a=3

b2=1b=1

Compute the vertices.

The vertices is said to be ((h±a),k) .

Substitute the value 3 for a .

Then, the vertices of the ellipse is (1±3,2) .

Compute the value of c using the equation below.

c2=a2b2

Substitute the value 3 for a and 2 for b

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