Chapter 10, Problem 10RE

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

# Identifying a Conic In Exercises 7-14, identify the conic, analyze the equation (center, radius, vertices, foci, eccentricity, directrix, and asymptotes, if possible), and sketch its graph. Use a graphing utility to confirm your results. 5 x 2 + y 2 − 20 x + 19 = 0

To determine
The type of conic by analyzing the equation (center, radius, vertices, foci, eccentricity, directrix, and asymptotes, if possible), and sketching its graph. Using a graphing utility confirm your results.

Explanation

Given:

The equation is 5x2+y2âˆ’20x+19=0.

Explanation:

Consider the equation 5x2+y2âˆ’20x+19=0.

Here, x and y both have squared terms and their coefficients are positive. Thus, this equation represents an ellipse.

To find the center, write this equation in the form of the standard equation of an ellipse which is given by:

(xâˆ’h)2a2+(yâˆ’k)2b2=1

Rearrange the equation as:

5x2+y2âˆ’20x=âˆ’19

Regroup the terms as:

5x2âˆ’20x+y2=âˆ’19

Factoring the coefficients of square terms:

5(x2âˆ’4x)+y2=âˆ’19

Divide both sides by 5:

(x2âˆ’4x)+15y2=âˆ’195

Use the completing squares method:

(x2âˆ’4x+22)+15y2=âˆ’195+22

That is,

(xâˆ’2)2+15y2=âˆ’195+4

That is,

(xâˆ’2)2+15y2=15

Divide the above equation by 15.

(xâˆ’2)215+y2=1

Rearrange in standard form as:

(xâˆ’2)2(15)2+(yâˆ’0)2(1)2=1

Comparing this with the standard equation of an ellipse with center (h,k):

(xâˆ’h)2a2+(yâˆ’k)2b2=1

So, h=2, k=0

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