"Sketch the ellipse. Find the coordinates of its vertices and foci.
34. x^2 + 25y^2 - 6x - 100y + 84 = 0
(Hint: Complete the squares in x and y. Begin by writing the equation in this form: (x^2 - 6x + ___) + 25(y^2 - 4y + ___) = -84.)"
==> I used the hint that came with the problem, but as I tried to solve the equation into the general form for an ellipse, I was getting negative denominators, which would make a and b contain imaginary numbers. Does anyone know how to solve this? Any help is greatly appreciated! :)
x^2 + 25y^2 - 6x - 100y + 84 = 0
x^ - 6x + 9 + 25(y^2 - 4y + 4) = -84 + 9 + 100
notice in the y expression, I added 4 to the inside, but that was multiplied by 25, so I had to add 100 on the right side.
(x-3)^2 + 25(y-2)^2 = 25
(x-3)^2/25 + (y-2)^2 = 1
so a^2 = 25, and b^2 = 1
take it from there
Ohhhhh!!!!! This totally makes sense. I forgot to multiply the 4 by 25 before bringing it over to the other side. Thanks so much!! :D
To sketch the ellipse and find the coordinates of its vertices and foci, let's start by completing the squares in the given equation.
Starting with the equation:
x^2 + 25y^2 - 6x - 100y + 84 = 0
Rearrange the terms by grouping the variables:
(x^2 - 6x) + 25(y^2 - 4y) = -84
To complete the square for x, take half of the coefficient of x (-6) and square it (-6/2)^2 = 9. Add 9 to both sides:
(x^2 - 6x + 9) + 25(y^2 - 4y) = -84 + 9
Similarly, to complete the square for y, take half of the coefficient of y (-4) and square it (-4/2)^2 = 4. Add 4 to both sides:
(x^2 - 6x + 9) + 25(y^2 - 4y + 4) = -84 + 9 + 25(4)
Simplify:
(x - 3)^2 + 25(y - 2)^2 = -84 + 9 + 100
(x - 3)^2 + 25(y - 2)^2 = 25
Now divide both sides by the constant on the right side of the equation to normalize the equation:
[(x - 3)^2]/25 + [(y - 2)^2]/1 = 1
The equation is now in the standard form of an ellipse:
[(x - h)^2]/a^2 + [(y - k)^2]/b^2 = 1
Where the center is represented by the coordinates (h, k), and a and b are the semi-major and semi-minor axes lengths.
In our case:
Center: (3, 2)
a = √25 = 5
b = √1 = 1
So the center of the ellipse is (3, 2), the semi-major axis length is 5, and the semi-minor axis length is 1.
To find the coordinates of the vertices, recall that the vertices lie along the major axis, which is parallel to the x-axis. This means the y-coordinate remains constant at k = 2, and the x-coordinate changes by a.
Vertices:
(3 + 5, 2) = (8, 2) and (3 - 5, 2) = (-2, 2)
Now, to find the coordinates of the foci, use the equation:
c^2 = a^2 - b^2
In our case:
c^2 = 5^2 - 1^2
c^2 = 25 - 1
c^2 = 24
c = √24
The foci lie along the major axis, at a distance c from the center.
Foci:
(3 + √24, 2) ≈ (8.90, 2) and (3 - √24, 2) ≈ (-2.90, 2)
Now you have the sketch of the ellipse and the coordinates of its vertices and foci.