   Chapter 10, Problem 46RE

Chapter
Section
Textbook Problem

# Find the foci and vertices and sketch the graph.46. 4x2 − y2 = 16

To determine

To find: The foci and vertices and sketch the graph of 4x2y2=16.

Explanation

Calculation:

Rewrite the equation 4x2y2=16 to a standard form.

4x2y2=16 (1)

Divide the above equation by 16 on both sides.

4x216y216=1616x24y216=1

(x0)222(y0)242=1 (2)

The standard equation of the hyperbola is as follows.

(xh)2a2(yk)2b2=1 (3)

Here h,k is the center of the ellipse, a is x-intercept, and b is y-intercept.

Compare the equation (2) and (3).

h,k=(0,0)

Here, in the given equation 9 is greater than 8. So, 9 is taken as. a2

a2=4;a=±2

b2=16;b=±4

Find the distance of the foci from the center point c is c2=a2+b2.

Substitute a=±2 and b=±4 in c2=a2+b2,

c=4+16c=20

That is, c=±25 (4)

Determine the foci.

The foci are located at points (0±c,0)=(±25,0).

The vertices are at points (h±a,k)=(0±2,0)=(±2,0).

Thus, the foci are at (±25,0)_. The vertices are at (±2,0)_.

The equation of the asymptotes is y=±bax

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