Identify the shape of the following equation: 3x^2 + 4y^2 -12x-36 = 0?

Question

you can rearrange the numbers anyway you want. maybe you like it better as: 3x^2 -12x + 4y^2 = 36

like is it a circle? hyperbola? how do you tell??

Answer 1

We'll complete the square:

3x^2 - 12x + 4y^2 = 36
x^2 - 4x + (4/3)y^2 = 12
(x^2 - 4x + 4) + (4/3)y^2 = 12 + 4 = 16
(x-2)^2 + y^2/(3/4) = 16

(If the x and y terms had the same coefficients, it would be a circle.)

[(x-2)^2]/16 + (y^2)/12 = 1

This is an ellipse centered at (2,0) with a horizontal major axis of 4 and a vertical minor axis of sqrt(12) = 2 sqrt(3).

To check, evaluate at these points: (-2,0), (6,0), (2, 2 sqrt 3), (2, -2 sqrt 3): f(-2,0) = 1 + 0 = 1; f(6,0) = 1 + 0 = 1; f(2,2 sqrt 3) = 0 + 1 = 1; f(2, -2 sqrt 3) = 0 + 1 = 1.

It all checks out.

Answer 2

This is a circle. Circles are in the form of (x - h)^2 + (y - k)^2 = r^2.
X = 3x^2
H = 12x
Y = 4y^2
K = 0
your radius would be the square root of 36, which is 6.

Usually, when you look at the equation, your squared variable is the X or Y. For example X^2 is the X in the circle equation, and the lone variable is either K or H. Your constant would be the radius squared.

I don't know if this makes any sense to you. I hope it helped and didn't confuse you more.

Answer 3

It is an ellipse.
Why? I looked at the coefficients of the terms x^2 and y^2.
The coefficient of x^2 is 3.
The coefficient of y^2 is 4.
Both numbers are positive and unequal. Therefore, it is an ellipse.
If the numbers were positive and equal, then it would be a circle.
If the numbers had opposite signs, then it would be a hyperbola.

Answer 4

3x^2 + 4y^2 - 12x - 36 = 0
3x^2 - 12x + 4y^2 - 36 = 0
(3x^2 - 12x) + 4y^2 - 36 = 0
3(x^2 - 4x) + 4y^2 - 36 = 0
3(x^2 - 4x + 4 - 4) + 4y^2 - 36 = 0
3((x - 2)^2 - 4) + 4y^2 - 36 = 0
3(x - 2)^2 - 12 + 4y^2 - 36 = 0
3(x - 2)^2 + 4y^2 - 48 = 0
3(x - 2)^2 + 4y^2 = 48
(((x - 2)^2)/4) + ((y^2)/3) = 4

ANS : Ellipse

For a graph, go to www.quickmath.com

Answer 5

You can tell by the signs...

general equation for a circle is (x - h)^2 + (y - k)^2 = r^2
general equation for an elipse is (x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
general equation for a hyperbola is x^2/a&^2 - y ^2/b^2 = 1
general equation for a parabola is (y - k)2 = 4a(x - h)

I think that is an elipse

If you need more help check here:
http://www.analyzemath.com/EllipseEq/EllipseEq.html

Answer 6

complete the squares to see what type of formula you have.

Answer 7

It is an ellipsis.