## Archive for the ‘Conic Sections’ Category

## 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; x^{2} + y^{2} + ax +by +c = 0

So the equation is

x^{2} + y^{2} + 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} = r^{2}

So the equation is

=> (x +3)^{2} + (y – 4)^{2} = 4^{2}

=> (x +3)^{2} + (y – 4)^{2} = 16

^{ }

## Problem 1122

Write the equation x^{2} + y^{2} -4x -6y +8 = 0 in the standard form.

**Solution:-**

Complete the square and write in standard form:

(x -h)^{2} + (y – k)^{2} = r^{2}

So the equation is

(x -2)^{2} + (y – 3)^{2} = 5

## Problem 1121

Write the equation in standard form.

x^{2} + y^{2} + 4y – 45 = 0

**Solution:-**

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Regroup the terms.

(x^{2}) + (y^{2} + 4y) – 45 = 0

Since x is square, there is nothing needed.

Now complete the square for y^{2} + 4y.

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

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

Now remove the subtract term from within the parentheses.

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

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

x^{2} + (y^{2} + 4y + 4 ) = 49

Now writ the equation in standard form.

(x -h)^{2} + (y – k)^{2} = r^{2}

(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^{2 }.

The radius of a circle is (4).

## Problem 1119

What is the center of the circle

(x + 3)^{2} + y^{2} = 4?

**Solution:-**

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

(x – h)^{2} + (y – k)^{2} = r^{2}

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

## Problem 1118

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

x^{2} + y^{2} + 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} = r^{2}

Regroup the terms.

(x^{2}) + (y^{2} + 4y) – 45 = 0

Since x is square, there is nothing needed.

Now complete the square for y^{2} + 4y.

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

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

Now remove the subtract term from within the parentheses.

x^{2} + (y^{2} + 4y + 4 – 4 ) – 45 = 0

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

x^{2} + (y^{2} + 4y + 4 ) = 49

Now writ the equation in standard form.

(x -h)^{2} + (y – k)^{2} = r^{2}

(x – 0)^{2} + (y + 2)^{2} = 49

h = 0

k = -2

The radius is 7.

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.

x^{2} + y^{2} + 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.

(x^{2} + 3x ) + (y^{2} – 5y) =

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

(x^{2} + 3x + ) + (y^{2} -5y) =

Now complete square for y^{2} – 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} = r^{2}

h = –

k =

The radius is .

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.

3x^{2} – 4x + 2 = 0

**Solution:-**

b^{2}– 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. |

b^{2}– 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. |

b^{2}– 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.

b^{2} – 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.