identify the conic section. Write in standard form and graph.
5x^2-6y^2-30x-12y+9=0
5x^2-6y^2-30x-12y+9=0
lets complete the square
5(x^2 - 6x + ___) -6(y^2 + 2y + ___) = -9
5(x^2 - 6x + 9) -6(y^2 + 2y + 1) = -9 + 45 - 6
5(x-3)^2 - 6(y+1)^2 = 30
divide both sides by 30
(x-3^2 /6 - (y+1)^2 /5 = 1
I will leave it up to you to identify the conic and its properties.
To identify the conic section and write it in standard form, we need to manipulate the given equation. Let's begin by rearranging the terms:
5x^2 - 6y^2 - 30x - 12y + 9 = 0
Rearranging the equation slightly:
5x^2 - 30x - 6y^2 - 12y + 9 = 0
Next, let's complete the square for both the x and y terms. We'll start with the x terms:
5x^2 - 30x = 6y^2 + 12y - 9
First, divide both sides by the coefficient of x^2 to make the leading coefficient 1:
x^2 - 6x = (6/5)y^2 + (12/5)y - (9/5)
To complete the square for the x-terms, we take half the coefficient of the x-term (-6/2 = -3) and square it (-3)^2 = 9:
x^2 - 6x + 9 = (6/5)y^2 + (12/5)y - (9/5) + 9
Simplifying:
(x - 3)^2 = (6/5)y^2 + (12/5)y + (36/5)
Now, let's complete the square for the y terms:
(x - 3)^2 = (6/5)y^2 + (12/5)y + (36/5)
First, divide both sides by the coefficient of y^2 to make the leading coefficient 1:
(x - 3)^2 / (6/5) = y^2 + (12/5)y + (36/5)
To complete the square for the y-terms, we take half the coefficient of the y-term (12/5/2 = 6/5) and square it:
(x - 3)^2 / (6/5) = (y + 6/5)^2 - (36/25) + (36/5)
Simplifying:
(x - 3)^2 / (6/5) = (y + 6/5)^2 + 36/25
Finally, we can write the equation in standard form:
(x - 3)^2 / (6/5) - (y + 6/5)^2 / (36/25) = 1
Based on the equation, the conic section is a hyperbola.
To graph the hyperbola, we can start by plotting the center point, which is (h, k) = (3, -6/5). Next, we can draw the asymptotes, which are lines that pass through the center and intersect the vertices of the hyperbola. The general equation for the asymptotes of a hyperbola with center (h, k) is given by:
y - k = ±(b/a)(x - h)
where a and b are the lengths of the conjugate and transverse axes, respectively.
In this case, since the hyperbola equation is in standard form:
(x - 3)^2 / (6/5) - (y + 6/5)^2 / (36/25) = 1
We can determine that the length of the conjugate axis is 2b = 2 * sqrt(a^2 + c^2), where c is the distance from the center to the foci. The length of the transverse axis is 2a. Based on these calculations, we can plot the vertices, foci, and asymptotes to form the graph of the hyperbola.