What conic section is represented by the equation 3x^2+2y^2-24x+30=0 ? Explain.
wrong again!
The standard form for an ellipse is
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
3x^2 + 2y^2 - 24x + 30 = 0
3(x^2-8x+16) + 2(y^2) = -30 + 48
3(x-4)^2 + 2y^2 = 18
(x-4)^2/6 + y^2/9 = 1
so a = 3, b=√6, c = √3
so it is an ellipse with center at (4,0)
and major axis along the y-axis.
with vertices at (0,3) and covertices at (4±√6,0)
and foci at (4,±√3)
Apologies for the incorrect response. You are correct, the standard form for an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1. Thank you for providing the correct standard form and explaining the solution.
The given equation 3x^2 + 2y^2 - 24x + 30 = 0 can be rewritten as 3(x^2-8x+16) + 2y^2 = -30 + 48. Further simplification leads to 3(x-4)^2 + 2y^2 = 18.
By comparing this equation to the standard form (x-h)^2/a^2 + (y-k)^2/b^2 = 1, we can conclude that the center of the ellipse is at the point (4, 0), with major axis along the y-axis. The equation also tells us that a^2 = 6 and b^2 = 9, so a = sqrt(6) and b = 3.
Therefore, the ellipse has a major axis of length 2a = 2sqrt(6) and a minor axis of length 2b = 6. The vertices of the ellipse are located at (4, 3) and (4, -3), and the co-vertices are located at (4 + sqrt(6), 0) and (4 - sqrt(6), 0). The foci of the ellipse are located at (4, sqrt(3)) and (4, -sqrt(3)).
The given equation 3x^2 + 2y^2 - 24x + 30 = 0 represents an ellipse.
To determine the conic section, we need to rewrite the equation in a specific form. An equation of an ellipse can be represented in the general form:
A(x-h)^2 + B(y-k)^2 = 1
Comparing the given equation with the general form of an ellipse equation, we observe that the terms with variables x and y are squared (x^2 and y^2), therefore, it must be an ellipse.
To convert the equation into the standard form of an ellipse, we need to complete the square for both the x and y terms. Let's start with the x terms:
3x^2 - 24x = 0
We divide the whole equation by 3 to simplify it:
x^2 - 8x = 0
Now, we complete the square for the x terms:
x^2 - 8x + 16 = 16
Moving to the y terms:
2y^2 = 24 - 3x^2
Dividing the whole equation by 2:
y^2 = 12 - (3/2)x^2
Now, let's make the coefficient of x^2 equal to 1 by dividing the equation by 12:
(1/12)y^2 = 1 - (1/4)x^2
Comparing it to:
A(x-h)^2 + B(y-k)^2 = 1
We can see that h = 2 (since -8/2 = -4) and k = 0 (since we don't have any constant term). Also, A = 4 (since 1/4 = 1/2^2) and B = 12 (since -1/12 = -1/2^2).
Thus, the standard form of the given equation is:
4(x-2)^2 + 12y^2 = 16
This is the equation of an ellipse with center (2, 0) and major axis along the x-axis.