## Problem 1143

Find the eccentricity of the ellipse

Solution:-

eccentricity of the ellipse = c/a

c = √(a^2 – b^2)

c = √(25-5)

c= √(20)

c = 2√(5)

eccentricity = (2√5)/5

## Problem 1124

Write the general form equation of (x + 3)2 + (y + 6)2 = 3.

Solution:-

FOIL the binomials and write in general

Form; x2 + y2 + ax +by +c = 0

So the equation is

x2 + y2 + 6x +12y + 42 = 0

## Problem 1123

Write the standard form equation of a circle with a center at (-3, 4) and a radius of 4.

Solution:-

Substitute the coordinates and write in standard form:

(x -h)2 + (y – k)2 = r2

So the equation is

=> (x +3)2 + (y – 4)2 = 42

=> (x +3)2 + (y – 4)2 = 16

## Problem 1122

Write the equation x2 + y2 -4x -6y +8 = 0 in the standard form.

Solution:-

Complete the square and write in standard form:

(x -h)2 + (y – k)2 = r2

So the equation is

(x -2)2 + (y – 3)2 = 5

## Problem 1121

Write the equation in standard form.

x2 + y2 + 4y – 45 = 0

Solution:-

Complete the square and then write the equation in standard form.

(x – h)2 + (y – k)2 = r2

Regroup the terms.

(x2) + (y2 + 4y) – 45 = 0

Since x is  square, there is nothing needed.

Now complete the square for y2 + 4y.

To complete the square, add and subtract 4 inside the second set of parentheses.

x2 + (y2 + 4y + 4 – 4 ) – 45 = 0

Now remove the subtract term from within the parentheses.

x2 + (y2 + 4y + 4 – 4 ) – 45 = 0

Use the addition principle to get the constant terms on the right side of the equation.

x2 + (y2 + 4y + 4  ) = 49

Now writ the equation in standard form.

(x -h)2 + (y – k)2 = r2

(x – 0)2 + (y + 2)2 = 49

## Problem 1120

What is the radius  of the circle

(x + 16)2 + (y – 8)2 = 16 ?

Solution:-

The radius of a circle is the r-value from the equation (x – h)2 + (y – k)2 = r.

The radius of a circle is (4).

## Problem 1119

What is the center of the circle

(x + 3)2 + y2 = 4?

Solution:-

The center of a circle is  (h, k ) from the equation

(x – h)2 + (y – k)2 = r2

The center of the circle is (-3, 0).

## Problem 1118

Find the center and radius of the circle. Then graph the circle.

x2 + y2 + 4y – 45 = 0

Solution:-

To find the center and the radius, you must complete the square and then write the equation in standard form.

(x – h)2 + (y – k)2 = r2

Regroup the terms.

(x2) + (y2 + 4y) – 45 = 0

Since x is  square, there is nothing needed.

Now complete the square for y2 + 4y.

To complete the square, add and subtract 4 inside the second set of parentheses.

x2 + (y2 + 4y + 4 – 4 ) – 45 = 0

Now remove the subtract term from within the parentheses.

x2 + (y2 + 4y + 4 – 4 ) – 45 = 0

Use the addition principle to get the constant terms on the right side of the equation.

x2 + (y2 + 4y + 4  ) = 49

Now writ the equation in standard form.

(x -h)2 + (y – k)2 = r2

(x – 0)2 + (y + 2)2 = 49

h = 0

k = -2

To sum up, the center is (0, -2) and the radius is 7.

To graph the circle, first plot the center (0,-2).

To complete the circle, use the radius to find all the point equidistant from the center.

## Problem 1117

Find the center and radius of the circle. Then graph the circle.

x2 + y2 + 3x – 5y – = 0

Solution:-

To find the center and the radius, complete the square and then write the equation in standard form.

(x – h)2 + (y – k)2 = r

Regroup the terms.

(x2 + 3x ) + (y2 – 5y) =

First complete the square, add to both sides of the equation.

(x2 + 3x + ) + (y2 -5y) =

Now complete square for y2 – 5y.

To complete the square, add to both sides of the equation.

(x + 3x + ) + (y – 5y + )

=

Simplify the right-hand side.

(x + 3x + ) + (y – 5y + )=

Now write the equation in standard form.

(x – h)2 + (y – k)2 = r2

h = –

k =

To sum up, the center is () and the radius is .

To graph the circle, first plot the center (-).

To complete the circle, use the radius to find all the points equidistant from the center.

## Problem 1086

Compute the discriminant. Then determine the number and type of solution for the given equation.

3x2 – 4x + 2 = 0

Solution:-

 b2– 4ac > 0 Two unequal real solutions; if a, b, and c are rational numbers and the discriminant is a perfect square, the solution are rational. If the discriminant is not a perfect square, the solution are irrational. b2– 4ac = 0 One solution (a repeated solution) that is a real number; if a, b, and c are rational number, the repeated solution is also a rational number. b2– 4ac < 0 No real solution; two complex imaginary solution; The solution are complex conjugates.

Using a = 3, b = -4, and c = 2, we evaluate the discriminate.

b2 –  4ac = (-4)2 – 4 (3)(2) = -8

since b – 4ac < 0 there is no real solution to the quadratic equation. The solutions are complex conjugates.