Archive for the ‘Conic Sections’ Category

Problem 1143

Find the eccentricity of the ellipse   \frac{x^{2}}{25}+\frac{(y-7)^{2}}{5}=1

 

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

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).

 

c3

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

 

c4

 

 

Problem 1117

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

x2 + y2 + 3x – 5y – \frac{91}{4} = 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) = \frac{91}{4}

First complete the square, add \frac{9}{4} to both sides of the equation.

(x2 + 3x + \frac{9}{4}) + (y2 -5y) = \frac{91}{4} + \frac{9}{4}

Now complete square for y2 – 5y.

To complete the square, add \frac{25}{4} to both sides of the equation.

(x + 3x + \frac{9}{4}) + (y – 5y + \frac{25}{4})

= \frac{91}{4}+\frac{9}{4}+\frac{25}{4}

Simplify the right-hand side.

(x + 3x + \frac{9}{4}) + (y – 5y + \frac{25}{4})=\frac{125}{4}

Now write the equation in standard form.

(x – h)2 + (y – k)2 = r2

(x+\frac{3}{2})^{2}+(y-\frac{5}{2})^{2}=\frac{125}{4}

h = –\frac{3}{2}

k = \frac{5}{2}

The radius is \frac{5\sqrt{5}}{2}.

To sum up, the center is (-\frac{3}{2},\frac{5}{2}) and the radius is \frac{5\sqrt{5}}{2}.

To graph the circle, first plot the center (-\frac{3}{2},\frac{5}{2}).

c1

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

 

c2

 

 

 

 

 

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.