# Precal help questions?

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)

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.

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 .

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

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.

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)].

The conic equation

1) To identify the conic section

General form of a conic equation inthe form

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

is ellipse.

2) To find the ellipse in standard form.

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.

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.

Compare it to standard form of ellipse

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.

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

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.