Anyone who is an expert with conics and completing squares please feel free to help me, i need the help badly?

Question

Completing the square help?

how do i complete the square for the following? I can't figure any of these out, please show me all the steps!
1. (9x^2)+(25y^2) = 225
2. (9x^2)-(2y^2)+18 = 0
3. (y^2)-(6y)+(16x)+25 = 0
4. (2x^2)+(2y^2)-(10x)-18y = 1

Answer 1

3 and 4 are the only ones that require completion of a square. See my answer here:
http://answers.yahoo.com/question/index;_ylt=Ap_MClwALAqFXwxyMyEZhj_sy6IX?qid=20070826211926AAf4k4l

for why the other two do not require completion of a square. Completing the square involves subtracting / adding something to an equation (which doesn't change it). For example:

x = y
x + 2 - 2 = y

The second equation is the same as the first, but if u=x+2, we can then say:
u - 2 = y

See:
http://en.wikipedia.org/wiki/Completing_the_square
for more on completing the square.

#3
We need something like:
(y + a)² = y² - 6y + a²
Expand the left hand side:
y² + 2ay + a² = y² - 6y + a²
That means a = -3, so we do the following:

y² - 6y + 16x + 25 = 0
y² - 6y + 9 - 9 + 16x + 25 = 0
(y² - 6y + 9) - 9 + 16x + 25 = 0
(y - 3)² - 9 + 16x + 25 = 0
(y - 3)² + 16x + 25 - 9 = 0
(y - 3)² + 16x + 16 = 0

#4
Here we have to complete two squares:
2(x+a)² = 2x² - 10x + a²
Leads us to:
a = -5/2
2(y+b)² = 2y² - 18y + b²
Leads us to:
b = -9/2

And we then get:

2x² + 2y² - 10x -18y = 1
2x² - 10x + 2y² -18y = 1
2x² - 10x + 25/2 - 25/2 + 2y² -18y = 1
2(x² - 5x + 25/4) -25/2 + 2y² -18y = 1
2(x - 5/2)² - 25/2 + 2y² -18y = 1
2(x - 5/2)² - 25/2 + 2y² -18y +81/2 - 81/2 = 1
2(x - 5/2)² - 25/2 + 2(y² -18y +81/4) - 81/2 = 1
2(x - 5/2)² - 25/2 + 2(y - 9/2)² - 81/2 = 1
2(x - 5/2)² + 2(y - 9/2)² - 81/2 - 25/2 = 1
2(x - 5/2)² + 2(y - 9/2)² - 106/2 = 1
2(x - 5/2)² + 2(y - 9/2)² - 53 = 1

Answer 2

1. This is an ellipse. The standard form of an ellipse is:

((x - h)/rx)^2 + ((y - k)/ry)^2 = 1

Where:
rx is the length of the radius in the+/- x-direction.
ry is the length of the radius in the+/- y-direction.

h and k are the offsets from the origin on the ellipse and in this case are 0 since we just have terms of x^2 and y^2.

Divide by 9*25 (equal to 225), this will remove the 9 and 25 in the numerators and, so the problem can be solved easily, make the right hand side 1.

x^2/25 + y^2/9 = 1
(x/5)^2 + (y/3)^2 = 1 .... the standard form

2. This is a hyperbola (since the signs of the x^2 and y^2 terms are different, one positive and one negative). The standard form is:

((x - h)/rx)^2 - ((y - k)/ry)^2 = 1

Where:
rx is the distance from the center to the hyperbola's +/- x-direction vertex (or asymptote).
ry is the distance from the center to the hyperbola's+/- y-direction vertex (or asymptote).

Again h,k is the center and these are both 0 since there are only terms in x^2 and y^2 (if these are non-zero you would also have terms in x and/or y.

(9x^2) - (2y^2) + 18 = 0
18 = -(9x^2) + (2y^2)

The hyperbola opens in the +/-y direction since the sign in front of the y^2 term is positive (If you set y = 0 there is no real solution so the hyperbola can not pass through the x-axis, if you set x = 0 then the points are 18 = 2y^2 and y = +/-3 - OK?).

Divide each side by 2*9 (equal to 18) to get rid of the numbers in the numerators and to, coincidentally make the left hand side equal to 1.

1 = -x^2/2 + y^2/9
1 = -(x/SQRT(2))^2 + (y/9)^2

3. This is a parabola since there is no x^2, only a y^2. This must be written in "vertex form" to complete the squares (there is no square term for x since it is not squared):

x = a(y – h)^2 + k

(y^2) - (6y) + (16x) + 25 = 0

Rearrange (put the x on one side):
16x = -y^2 + 6y - 25
16x = -(y^2 - 6y) - 25

In general you can write:
(y - h)^2 = y^2 - 2h + h^2

In this case 2h = 6 and h = 3, h^2 = 9
So you must include a +9 inside the parenthesis to give a term (y - 3)^2 = y^2 - 6y + 9

16x = -(y^2 - 6y) - 25 = -(y^2 - 6y + 9) - 16
16x = -(y - 3)^2 - 16
x = (-1/16)(y - 3)^2 - 1

4. This is a circle since the quadratic terms are 2x^2 and 2y^2 and they are added. It is offset from (0,0) since there are terms involving both x and y. The standard form of the circle is:

(x - h)^2 + (y - k)^2 = r^2

Separate and join the x and y terms:

2x^2 - 10x + 2y^2 - 18y = 1

Divide by 2:

(x^2 - 5x) + (y^2 - 9y) = 1/2

as above you want terms of the form: (x - a)^2 and (y - b)^2
using this sort of thing for both:

(y - h)^2 = y^2 - 2h + h^2

a = 5/2 and b = 9/2

(x^2 - 5x) = (x - 5/2)^2 - (5/2)^2 = x^2 - 5x + (5/2)^2 - (5/2)^2
and: (x^2 - 5x) = (x^2 - 5x)

Likewise for y: (y^2 - 9y) = (y - 9/2)^2 - (9/2)^2

So the equation can now be written as:
(x^2 - 5x) + (y^2 - 9y) = (x - 5/2)^2 - (5/2)^2 + (y - 9/2)^2 - (/9/2)^2 = 1/2
(x - 5/2)^2 + (y - 9/2)^2 = 1/2 + (5/2)^2 + (9/2)^2
(x - 5/2)^2 + (y - 9/2)^2 = 108/4 = 27 = (3SQRT(2))^2 = (4.24)^2

This is a circle with it's center at (5/2,9/2) and having a radius of 4.24