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Precal help questions?

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Given the equation x^2-2x+4y^2-16y+13=0

1. Identify the conic
2. Put the equation in standard from
3. Find coordinates of center ,vertices, foci, asymptoes
3. Sketch graph

Please explain how I can do this and what the correct answer would be :)

2. Given the complex number root3-3i

a. Find the trig form in the interval [o,2pi) and r>0.
b. using previous answer, find (root3-3i)^4 and write the answer in complex form (a+bi)
asked Apr 30, 2013 in PRECALCULUS by linda Scholar

6 Answers

0 votes

1.

x2 - 2x + 4y2 - 16y + 13 = 0

x2 - 2x + 1 + 4y2 - 16y + 12 = 0

x2 - 2x + 1 + 4y2 - 16y + 16 = 4

(x - 1)2 + (2y - 4)2 = 4

Divide each side by 4

(x - 1)2  / 4+ (2y - 4)2 / 4 = 1

Simplify

(x - 1)2  / 4 + (y - 2)2 / 1 = 1

This conic is a ellipse.

 

 

answered May 3, 2013 by diane Scholar
0 votes

2.

1.

x2 - 2x + 4y2 - 16y + 13 = 0

x2 - 2x + 1 + 4y2 - 16y + 12 = 0

x2 - 2x + 1 + 4y2 - 16y + 16 = 4

(x - 1)2 + (2y - 4)2 = 4

Divide each side by 4

(x - 1)2  / 4+ (2y - 4)2 / 4 = 1

Simplify

(x - 1)2  / 4 + (y - 2)2 / 1 = 1

This  is a ellipse standard form .

 

 

answered May 3, 2013 by diane Scholar
0 votes

The ellipse equation : (x - 1)2 / 4+ (y - 2)2 / 1 = 1

The center of the ellipse equation :(h , k) = (1 , 2)

a2 = 4 then a = ±2

b2 = 1 then b = ±1

c2 = a2 - b2 = 4 - 1 = 3

c = ±√(3)

If a>b then it is a horizontal axis

Substitute h = 1 , k = 2 and b = ±1 and a = ±2

The foci of an ellipse                   = (h , k ± c) = (1 , 2 ± √(3))

The vertices of anellipse              = (h , k ± b) = (1 , 2 ± 1) = (1 , 3) and (1 , 1)

The asymptotes of an ellipse       = y = ±(a / b)x = ± 2x

answered May 4, 2013 by diane Scholar
0 votes

2.

b.

(3 - 3i)4 = [(3 - 3i)2]2

Recall : Alzebra formula (a - b)2 = a2 - 2ab + b2

             = [9 - 18i + 9i2]2

Recall : complex number i , i2 = -1

             = [9 - 18i - 9]2

             = [0 - 18i]2

             = 324(i2)

Substitute i^2 = -1

             = 324(-1)

             = -324

We can rewrite the a + ib form

             = -324 + 0i.      

answered May 5, 2013 by diane Scholar
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2.The complex number root 3 - 3i

a.

Let z = x - iy = 3 - 3i

x = 3 , y = 3

|z| = r = √(x2 + y2) = sqrt(9 + 9) = √(18)

tanα = y / x = 3 / 3 = 1

α = pi / 4 , 3pi/ 4 interval[0, 2pi)

The trig form z = |z|(cosα - isinα)

Substitute r = |z| = √(18) and α = pi / 4 , 3pi / 4 in the above trig form

z = √(18)(cos(pi / 4) - isin(pi / 4))

or

z = √(18)[cos(3pi / 4) - i sin(3pi / 4)].

 

 

answered May 5, 2013 by diane Scholar
0 votes

The conic equation image

1) To identify the conic section

General form of a conic equation inthe form image

Both variables are squared and have the same sign, but they aren't multiplied by the same number, so this is an ellipse.

image is ellipse.

2) To find the ellipse in standard form.

image

image

To change the expressions (x 2- 2x) and (y 2 - 4y) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)² to each side of the equation.

image

image

image

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 4 to set it equal to 1.

image

image

Compare it to standard form of ellipse image

a 2 > b 2

If the larger denominator is under the "x " term, then the ellipse is horizontal.

3) Center (h, k ) = (1, 2)

a  = length of semi-major axis = 2

= length of semi-minor axis = 1

Vertices: (h + a, k ), (h - a, k )

= (1+2, 2) ,(1-2, 2)

Vertices are (3,2),(-1, 2)

is the distance from the center to each focus.

image

image

Foci: (h + c, k ), (h - c, k )

image

Foci (-0.73, 2), (2.73,2).

Ellipses do not have asymptotes.

By definition, an asymptote is a line that a graph approaches, but never intersects. It is a limit for the graph. Ellipses do not have such limits.

(h ,k ) = (1,2) , = 2 and b  = 1

The points for this ellipse are ,

Right most point (h +a , k )

Left most point (h - a , k )

Top most point (h , k + b )

Bottom most point (h , k - b )

Right most point (3, 2)

Left most point (-1, 2)

Top most point (1, 3)

Bottom most point (1, 1)

Graph

1. draw the coordinate plane.

2. Plot the center at (1, 2) and foci (-0.73, 2), (2.73,2).

3.Plot 4 points away from the center in the up, down, left and right direction.

4.Sketch the ellipse.

 

answered Jun 6, 2014 by david Expert

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