Find the center, vertices, foci, and eccentricity of the ellipse.
9x^2 + 4y^2 - 36x + 8y + 31 = 0
My answer was:
center=(2,1)
v=(-2,10)(-2,-10)
f=(-2,11)(-2,-11)
e=11/3
Where did I make a mistake?
9x^2 + 4y^2 - 36x + 8y + 31 = 0
9(x^2 - 4x + ...) + 4(y^2 + 2y + ...) = -31
9(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = -31+36+4
9(x-2)^2 + 4(y+1)^2 = 9
(x-2)^2 + (y+1)^2 /(9/4) = 1
my centre is (2,-1)
a^2 = 1
b^2 = 9/4) etc
have you found your error?
9^2-[(4.2x3.4)-9.28
To find the center, vertices, foci, and eccentricity of an ellipse, we need to rewrite the equation of the ellipse in standard form. The standard form of an ellipse is:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1
where (h, k) is the center of the ellipse, and a and b are the lengths of the semi-major and semi-minor axes, respectively.
Let's rewrite the given equation in standard form:
9x^2 + 4y^2 - 36x + 8y + 31 = 0
Start by grouping the x-terms and y-terms together:
9x^2 - 36x + 4y^2 + 8y + 31 = 0
To complete the square for x, divide the coefficient of x by 2, square it, and add it to both sides of the equation:
9(x^2 - 4x) + 4y^2 + 8y + 31 = 0
9(x^2 - 4x + 4) + 4y^2 + 8y + 31 = 9(4) + 4y^2 + 8y + 31
9(x - 2)^2 + 4y^2 + 8y + 31 = 36 + 4y^2 + 8y + 31
9(x - 2)^2 + 4(y^2 + 2y) + 31 = 36 + 4y^2 + 8y + 31
To complete the square for y, divide the coefficient of y by 2, square it, and add it inside the parentheses:
9(x - 2)^2 + 4(y^2 + 2y + 1) + 31 = 36 + 4y^2 + 8y + 31
9(x - 2)^2 + 4(y + 1)^2 + 31 = 36 + 4y^2 + 8y + 31
Now, rearrange the terms and simplify:
9(x - 2)^2 + 4(y + 1)^2 = 4y^2 + 8y + 36
Divide both sides of the equation by the constant term on the right-hand side to bring it to the value of 1:
[(x - 2)^2 / (4)] + [(y + 1)^2 / (9)] = (4y^2 + 8y + 36) / 36
[(x - 2)^2 / (4)] + [(y + 1)^2 / (9)] = (y^2 + 2y + 9) / 9
Now, we have the equation in standard form with a^2 = 4 and b^2 = 9. The center of the ellipse is (h, k), which corresponds to (2, -1).
To find the vertices, we need to find the values of a and b. The lengths of the semi-major and semi-minor axes are given by a and b, respectively. Therefore, a = 2 and b = 3.
The vertices are located at (h ± a, k). Therefore, the vertices are (2 ± 2, -1) = (0, -1) and (4, -1).
To find the foci of the ellipse, we need to find the value of c, which is given by c = sqrt(a^2 - b^2). In this case, c = sqrt(4 - 9) = sqrt(-5), which is not a real number. This indicates that the given equation does not represent an ellipse because the discriminant is negative. Therefore, there are no foci for this equation.
The eccentricity, e, of an ellipse is given by e = c / a. In this case, since c is not a real number, the eccentricity cannot be determined.
Thus, the corrected answer is:
Center: (2, -1)
Vertices: (0, -1) and (4, -1)
Foci: No foci exist
Eccentricity: Cannot be determined