ellipse has an equation of x^2+3y^2+16x+4=100y? standard form
try completing the squares:
x^2+16x + 3y^2-100y = -4
x^2+16x+64 + 3(y^2 - 100/3y + (50/3)^2) = -4 + 64 + 3*(50/3)^2
(x+8)^2 + (y - 50/3)^2 = 2680/3
unusual, but hey, it's just numbers, right?
But still, it's so odd that I suspect a typo. You sure it's not 4y^2? If so, I'm sure now you can see how to fix it.
oops. I botched it a bit, and lost the 3. We pick it up here:
(x+8)^2 + 3(y - 50/3)^2 = 2680/3
now get rid of that pesky 3:
(x+8)^2/3 + (y-50/3)^2 = 2680/9
To convert the equation of an ellipse to standard form, we need to complete the square for both the x and y terms. The standard form of an ellipse is:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
Where (h, k) represents the center of the ellipse, and a and b represent the lengths of the major and minor axes, respectively.
Let's rearrange the given equation step by step to get it in the standard form:
x^2 + 3y^2 + 16x + 4 = 100y
To begin, let's group the x terms and complete the square:
x^2 + 16x + 3y^2 + 4 = 100y
Now, let's factor out 3 from the y terms to complete the square:
x^2 + 16x + 4 + 3(y^2 - 100y) = 0
Next, we need to add and subtract the necessary terms to the equation to complete the square for both x and y:
x^2 + 16x + 64 + 4 + 3(y^2 - 100y + 2500) - 7500 = 0
Simplifying further:
(x + 8)^2 - 64 + 3(y - 50)^2 - 7500 = 0
(x + 8)^2 + 3(y - 50)^2 = 7564
Now, let's divide both sides of the equation by the constant term (7564) to normalize the equation:
((x + 8)^2) / 7564 + ((y - 50)^2) / 2521.33 = 1
Finally, we can write the equation of the ellipse in standard form:
((x + 8)^2) / 7564 + ((y - 50)^2) / 2521.33 = 1
So, the standard form of the given ellipse equation is ((x + 8)^2) / 7564 + ((y - 50)^2) / 2521.33 = 1.