For the equation 9x^2+4y^2+54x-8y+49=0 determine the center, vertices and foci?
rearrange stuff and complete the squares:
9x^2+54x + 4y^2-8y = -49
9(x^2+6x) + 4(y^2-2y) = -49
9(x^2+6x+9) + 4(y^2-2y+1) = -49 + 9*9 + 4*1
9(x+3)^2 + 4(y-1)^2 = 36
(x+3)^2/4 + (y-1)^2/9 = 1
Now answer the questions.
for this ellipse, the major axis is vertical.
a = 3
b = 2
The vertices are at (-3,1±3)
The co-vertices are at (-3±2,1)
Looks like it's time to review your ellipses.
how do you determine vertices?
To determine the center, vertices, and foci of the equation 9x^2 + 4y^2 + 54x - 8y + 49 = 0, we need to rewrite the equation in standard form, which is (x - h)^2/a^2 + (y - k)^2/b^2 = 1.
To do this, we'll complete the square for both the x and y terms. Let's start with the x terms:
9x^2 + 4y^2 + 54x - 8y + 49 = 0
Rearranging the equation:
9x^2 + 54x + 4y^2 - 8y = -49
Factoring out the common coefficients:
9(x^2 + 6x) + 4(y^2 - 2y) = -49
Now, to complete the square for the x terms, we take half of the coefficient of x (which is 6) and square it:
(6/2)^2 = 9
Adding 9 inside the parentheses, we get:
9(x^2 + 6x + 9) + 4(y^2 - 2y) = -49 + 9 * 9
Simplifying further:
9(x + 3)^2 + 4(y^2 - 2y) = -49 + 81
9(x + 3)^2 + 4(y^2 - 2y) = 32
Now, let's complete the square for the y terms:
9(x + 3)^2 + 4(y^2 - 2y) = 32
Taking half of the coefficient of y (which is -2) and squaring it:
(-2/2)^2 = 1
Adding 1 inside the parentheses, we get:
9(x + 3)^2 + 4(y^2 - 2y + 1) = 32 + 4 * 1
9(x + 3)^2 + 4(y - 1)^2 = 36
To put the equation in standard form, we need to divide both sides by 36:
[(x + 3)^2]/4 + [(y - 1)^2]/9 = 1
Now, we can see that the equation is in the form (x - h)^2/a^2 + (y - k)^2/b^2 = 1.
From this equation, we can identify the center as (-3, 1) because (h, k) represents the coordinates of the center.
The a^2 and b^2 values determine the length of the major and minor axes of the ellipse. In this case, a^2 = 4 and b^2 = 9. Taking the square root of these values, we have a = 2 and b = 3. Thus, the major axis is 2a = 4 units long, and the minor axis is 2b = 6 units long.
The vertices of the ellipse lie on the major axis and are a distance of a units away from the center. So, the vertices are (-3-2, 1) and (-3+2, 1), which simplifies to (-5, 1) and (-1, 1).
To find the foci of the ellipse, we will use the formula c = sqrt(a^2 - b^2), where c represents the distance between the center and each focus.
In this case, c = sqrt(4 - 9) = sqrt(-5). Since the square root of a negative number is imaginary, this means the ellipse has no real foci.
To summarize:
- The center of the ellipse is (-3, 1).
- The vertices are (-5, 1) and (-1, 1).
- The major axis has a length of 4 units (2a), while the minor axis has a length of 6 units (2b).
- The ellipse has no real foci.