Is the equation a line, parabola, ellipse, hyperbola, or circle?!?

Question

I can't figure it out for the life of me.

13x² + 13y² - 26x + 52y = -78 .

Is it a line, parabola, ellipse, hyperbola, or circle? how do I find out?

Answer 1

It is a circle
x^2+y^2 -2x+4y +6= 0
Center(1,-2) r^2= -1 its imaginary
(x-1)^2 +(y+2)^2 +1=0

Answer 2

13x^2 + 13y^2 - 26x + 52y = -78

1) its not a hyperbola because both signs on the x^2 and y^2 terms are the same (both positive. if they were both negative you could just multiply by -1).

2) not a line because of x^2 terms

3) not a parabola because it has both x^2 and y^2

so it could be an ellipse or a circle (which is a type of ellipse).

divide out 13

x^2 -2x + y^2 + 4y = 6

add 1 and 4 to both sides

x^2 - 2x + 1 + y^2 + 4y + 4 = 6 + 1 + 4 = 11
(x-1)^2 + (y + 2)^2 = 11
so it is a circle of radius sqrt(11), centered at (1,-2). (note that a circle is an ellipse with an eccentricity of 0). You can tell automatically that it will be a circle because the coefficient on the x^2 and y^2 terms is the same value so it will mean that a=b in your general form for an ellipse, which means your eccentricity will be 0 and therefore you will have a circle.

Answer 3

"I can't figure it out for the life of me". Well, yes, you *could*, but you'd need to know what you were doing!

You need to put it into standard form, and that takes some algebra. Gather together the similar variables

(13x^2 - 26x) + (13y^2+52y) = -78

Then factor out the 13's (and divide them out - they'll only get in the way).

(x^2 - 2x) + (y^2 + 4y) = -6

Now, complete each of the two squares.

(x^2 - 2x + 1) + (y^2 + 4y + 4) = -6 + 1 + 4 = -1

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

Congratulations, it isn't any of the three. You must have made an error in the equation you gave, it's what's called "degenerate", since there are no numbers that you can square and add up to get -1.

So, assuming it was +78, not -78, you get

(x-1)^2 + (y-2)^2 = 6 + 1 + 4 = 11

and it's a circle of radius sqrt(11).

A hyperbola would have one of the squared terms being subtracted, and an ellipse would have different factors for each one, like 2(x-1)^2 + 3(y-2)^2 = 11.

Now, start paying attention in class instead of begging for help online!

Carl

Answer 4

Rewrite the equation by

1) factoring out a 13 from the x^2 and x term, then complete the square inside your parenthesis and be sure to add the same amount to both sides of the equation when you do this. Remember whatever number completes the square is being multiplied by the 13 you factored out so add 13 times this number to the right side.

2) factor out a 13 from the y^2 and y term and complete the square as discussed above.

3) Now factor the two perfect squares on the left and add the three numbers on the right side.

Divide through by 13. Notice your perfect squares both have coefficients of 1 so this second degree equation is a circle.
However, if the number on the right side is negative it cannot be written as a positive radius squared. In this case you have what is called a degenerate circle.

First degree equations such as 2x + 5y = 1 are straight lines.

Second degree such as x^2 + (y-5)^2 = 3^2 are circles.

Second degree with differing positive coefficients such as

5x^2 + 9 y^2 = 25 are ellipses.

If the coefficients are of opposite siggs the graph is an hyperbola.

But if the constant on the right side is negative you have a degenerate conic.

Answer 5

hyperbola