Identifying Conic Sections...HELP!

A

ashleighw5678

New member
Joined
May 15, 2011
Messages
10
I have no idea how to do this.
Find the standard form of each equation by completing the square. Then identify and graph each conic.

x^2 + y^2 - 16x + 10y +53 = 0
 
Let's do this in pieces

Do you remember how to complete the square? It is probably the fundamental operation with respect to manipulating quadratics.

So what happens to the equation when you do that?
 
I don't remember how to complete the square very well. I know you add in a number on both sides... :?
 
You can complete the square by using the formula:

\(\displaystyle a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}\)

Take, for example, \(\displaystyle 2x^{2}+5x-7\)

We could add the square of half the coefficient of the linear term to both sides and so on.

But, we can enter a=2, b=5, c=-7 in the formula and go from there.

\(\displaystyle 2(x+\frac{5}{4})^{2}-\frac{81}{8}\)
 
ashleighw5678 said:
I have no idea how to do this.
Find the standard form of each equation by completing the square. Then identify and graph each conic.

x^2 + y^2 - 16x + 10y +53 = 0
Click to expand...
Let's resequence

x[sup:2egtasys]2[/sup:2egtasys] - 16x + y[sup:2egtasys]2[/sup:2egtasys] + 10y + 53 = 0.

Let's complete a square. The number that you add to both sides of the equation is zero, but on one side you add a non-zero number and subtract the same number.

0 = x[sup:2egtasys]2[/sup:2egtasys] - 16x + 64 - 64 + y[sup:2egtasys]2[/sup:2egtasys] + 10y + 53 = (x - 8)[sup:2egtasys]2[/sup:2egtasys] + y[sup:2egtasys]2[/sup:2egtasys] + 10y - 11.

Let's do it again.

0 = (x - 8)[sup:2egtasys]2[/sup:2egtasys] + y[sup:2egtasys]2[/sup:2egtasys] + 10y + 25 - 25 - 11 = (x - 8)[sup:2egtasys]2[/sup:2egtasys] + (y + 5)[sup:2egtasys]2[/sup:2egtasys] - 36.

OK That is completing the square. It's just basic algebra that you learned a while ago so it slipped your mind.

Does (x - 8)[sup:2egtasys]2[/sup:2egtasys] + (y + 5)[sup:2egtasys]2[/sup:2egtasys] - 36 = 0 suggest the canonical (standard) form of a particular conic section to you?
 
I am confused as to how you got the 64 in the first step.

Would the end result be a circle?

(Just a note, my teacher only taught us cirlce, ellipse, and hyperbola. He skipped parabolas.)
 
A perfect square

(a + b)[sup:1id1n3u7]2[/sup:1id1n3u7] = a[sup:1id1n3u7]2[/sup:1id1n3u7] + 2ab + b[sup:1id1n3u7]2[/sup:1id1n3u7].

In your case you have x[sup:1id1n3u7]2[/sup:1id1n3u7] -16x. Not yet a perfect square, but a = x and 2b = -16. So b = (-16 / 2) = - 8. And b[sup:1id1n3u7]2[/sup:1id1n3u7] = (- 8)[sup:1id1n3u7]2[/sup:1id1n3u7] = 64.

Make sense? You are finding the number to make your expression a perfect square.
 
I think so...I'm getting there. I feel ok on completing the square now, but I have no idea how to do this with conics. He didn't spend much time on this lesson.
 
Let's take this one step at a time. Do you see where the 25 came from when we completed the square for y[sup:611cr4yu]2[/sup:611cr4yu]?
 
Yes. We needed a number whose multiples added up to positive 10. So y^2 + 10y +25 would equal (y+5) (y+5) right?
 
Four...Circle, Ellipse, Hyperbola, and Parabola. We only studied Circle, Ellipse, and Hyperbola in depth.
 
Warning

When threads get long, they skip to a new page. There is a tiny icon that tells you that there is more than one page. Keep an eye out for it.

Yes there are four kinds, but it is sometimes more convenient to treat the circle as a special case of the ellipse.

For each conic there is a standard or canonical equation that describes the conic under SOME co-ordinate system.

For an ellipse that standard equation is (u / a)[sup:7agl5zv7]2[/sup:7agl5zv7] + (v / b)[sup:7agl5zv7]2[/sup:7agl5zv7] = 1.

In the special case of the circle, a = b so the standard equation reduces to (u[sup:7agl5zv7]2[/sup:7agl5zv7] + v[sup:7agl5zv7]2[/sup:7agl5zv7]) / a[sup:7agl5zv7]2[/sup:7agl5zv7] = 1.

So, in your expression the PLUS sign tells you that you are dealing with an ellipse or circle.
 
What does the u and v mean? Are you referring to the center as (U, V)? We learned it as (H,K) as used in these equations...

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

Ellipse: x^2/a^2 + y^2/b^2 = 1
 
ashleighw5678 said:
What does the u and v mean? Are you referring to the center as (U, V)? We learned it as (H,K) as used in these equations...

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

Ellipse: x^2/a^2 + y^2/b^2 = 1
Click to expand...

Great. Let's use the letters that you are used to.

You have (x - 8)[sup:fglpgx58]2[/sup:fglpgx58] + (y + 5)[sup:fglpgx58]2[/sup:fglpgx58] - 36 = 0.

Notice that both your standard ellipse equation and your standard circle equation have 2 terms on the left joined by a plus sign and a positive number on the right. Can we take the expression you have and make it look more like one of those standard forms?
 
You have (x - 8)[sup:n2uh3vcb]2[/sup:n2uh3vcb] + (y + 5)[sup:n2uh3vcb]2[/sup:n2uh3vcb] - 36 = 0.

Notice that both your standard ellipse equation and your standard circle equation have 2 terms on the left joined by a plus sign and a positive number on the right. Can we take the expression you have and make it look more like one of those standard forms?
Click to expand...

Ok, so would the radius be 6 and therefore the equation would be

(x-8)[sup:n2uh3vcb]2[/sup:n2uh3vcb] + (y+5)[sup:n2uh3vcb]2[/sup:n2uh3vcb] = 36

?
 
ashleighw5678 said:
Ok, so would the radius be 6 and therefore the equation would be

(x-8)[sup:311snwtj]2[/sup:311snwtj] + (y+5)[sup:311snwtj]2[/sup:311snwtj] = 36

?
Click to expand...

Yep, you have your equation in the form recommended by your teacher. You know what kind of curve to draw. You know the radius. Do you know where the center is?

I suspect your other problems will follow this format.
Complete the square. Manipulate the equation into a standard form. Draw the graph based on the equation in standard form.

See. That was not really that hard.
 
You must log in or register to reply here.
Top