help me solve this conic:
12x^2+y^2+6x-9=0
not sure what you mean by "solve this conic"
You want to know what kind of conic? ellipse
to find out its properties you have to change it into standard form (x-h)^2/a^2 + (y-k)^2/b^2 = 1
I would first divide by 12
x^2 + y^2/12 + x/2 = 9/12
complete the square for the x terms, adding the same quantity to both sides
x^2 + x/2 + 1/16 + y^2/12 = 3/4 + 1/16
(X + 1/4)^2 + Y^2/12 = 13/16
Divide each term by 13/16
(x + 1/4)^2/(13/16) + y^2/(26/16) = 1
so we have an ellipse with centre at (-1/4,0), with a = √13 /4 and b = √26 / 4
To solve the given conic, which is in the form 12x^2 + y^2 + 6x - 9 = 0, we want to determine the type and properties of the conic.
To begin, we need to transform the equation into standard form for an ellipse, which is given by (x-h)^2/a^2 + (y-k)^2/b^2 = 1.
Here are the steps to solve the conic:
1. Divide the equation by 12 to obtain a leading coefficient of 1:
x^2/12 + y^2/12 + x/2 - 9/12 = 0
2. Rearrange the equation to group the x terms together and the constants on the other side:
x^2/12 + x/2 + y^2/12 = 9/12
3. Complete the square for the x terms by adding (1/2)^2 = 1/4 to both sides:
x^2/12 + x/2 + 1/4 + y^2/12 = 9/12 + 1/4
Simplifying the right side:
x^2/12 + x/2 + 1/4 + y^2/12 = 13/12
4. Factor and simplify the terms:
(x + 1/4)^2/12 + y^2/12 = 13/12
5. Divide each term by 13/12 to obtain a leading coefficient of 1 on the right side:
(x + 1/4)^2/(13/12) + y^2/12 = 1
Now we have the equation in standard form for an ellipse. By comparing it to (x-h)^2/a^2 + (y-k)^2/b^2 = 1, we can determine the properties of the ellipse.
The center of the ellipse is given by the coordinates (h, k) = (-1/4, 0).
The semi-major axis, a, is equal to the square root of the coefficient in the denominator of the x-term, which is √13/4.
The semi-minor axis, b, is equal to the square root of the coefficient in the denominator of the y-term, which is √26/4.
Therefore, the given conic is an ellipse with a center at (-1/4, 0), a semi-major axis of √13/4, and a semi-minor axis of √26/4.