Hey, I got a few ellipse problems I'm havin trouble with. Any help appreciated.
Write in standard form, showing work. (standard form is (x^2 / a^2) + (y^2 / b^2) = 1, right?)
1. 16x^2 + 25y^2 - 64x - 100y + 564 = 0
2. 5x^2 + 3y^2 - 40x - 36y + 188 = 0
3. 3x^2 + 4y^2 - 12x + 16y + 16 = 0
Find the equation of an ellipse that has foci (0, +-2) and the endpoints of the minor axes are (+-5,0). Write in standard form, showing work.
the standard form of an ellipse with center (h,k) is:
[(x - h)^2]/a^2 + [(y - k)^2]/b^2 = 1
the formula you wrote is the standard form of an ellipse with center the origin.
starting from the equations you have given, we can get to the standard form by completing the square as you saw TPH did. however, he ended up with a negative number for the constant, so should he try to get 1 on one side, he would have to multiply by a negative number and the result wouldn't be the equation of an ellipse. check your questions again.
we see that the minor axis lies along the x-axis. the center is halfway between the endpoints of the axis. so here, the center is the origin (0,0).
the equation of an ellipse with center (0,0), and foci (0, +/- c) is:
x^2/b^2 + y^2/a^2 = 1
the minor axis is 2b = 10 (since the minor axis is 10 units long)
=> b = 5
the foci are (0, +/- 2), where c = +/-2.
now c^2 = a^2 - b^2
=> 4 = a^2 - 25
=> a^2 = 29
so our ellipse is:
x^2/25 + y^2/29 = 1
since we're apparently having problems with the first, let's try to find the equation for the other 2.
5x^2 + 3y^2 - 40x - 36y + 188 = 0
we proceed by completing the square:
5x^2 - 40x + 3y^2 - 36y = -188 ...........grouped terms with x's and y's togaether
=> 5(x^2 - 8x) + 3(y^2 - 12y) = -188
=> 5(x^2 - 8x + (-4)^2 - (-4)^2) + 3(y^2 - 12y + (-6)^2 - (-6)^2) = -188
=> 5[(x - 4)^2 - 16] + 3[(y - 6)^2 - 36] = -188
=> 5(x - 4)^2 - 80 + 3(y - 6)^2 - 108 = -188
=> 5(x - 4)^2 + 3(y - 6)^2 = 0
so we seem to be having a problem for this one as well.
3x^2 - 12x + 4y^2 + 16y + 16 = 03. 3x^2 + 4y^2 - 12x + 16y + 16 = 0
=> 3(x^2 - 4x) + 4(y^2 + 4y) + 16 = 0
=> 3(x^2 - 4x + (-2)^2 - (-2)^2) + 4(y^2 + 4y + (2)^2 - (2)^2) + 16 = 0
=> 3[(x - 2)^2 - 4] + 4[(y + 2)^2 - 4] + 16 = 0
=> 3(x - 2)^2 - 12 + 4(y + 2)^2 - 16 + 16 = 0
=> 3(x - 2)^2 + 4(y + 2)^2 = 12
=> [(x - 2)^2]/4 + [(y + 2)^2]/3 = 1
this one seems to work out ok.
I'll recheck my work for the first and second questions and see if i made a mistake