how i can isolve x^2 + 4y^2 = 1? it's an ellipse and i'm not quite sure how to treat the 4y^2.?

Question

thank you

Answer 1

What do you mean by how can you solve it?
It's an equation with two variables, there are an infinite amount of solutions for both...

Anyways...
x^2 + 4y^2 = 1
4y^2 = 1 - x^2
2y = +/- Sqrt(1 - x^2)
y = +/- Sqrt(1 - x^2)/2

Answer 2

x^2+4y^2=1, 4y^2 can be interpreted as 1/4^-1*y or
(1/2)^-2*y. By solving it I have no clue what you mean. This is a horizontal eclipse. The c or is the focus value because the eclipse is at the origin. It is sqrt of a^2-b^2

Answer 3

I think you want to know how to put this in standard form so you can figure out what a and b are?
Note that multiplying by four is the same as dividing by 1/4
x^2 + y^2/(1/4) = 1
So a^2 = 1 and b^2 = 1/4 or whatever your book uses in the std equation: x^2/a^2 + y^2/b^2 = 1
So the major vertices are at (-1,0), (1,0) and the endpoints are at (0, 1/2), (0,-1/2).

Answer 4

I suspect what you are trying to do is put it in a form where a and b can be more readily determined.

x² + 4y² = 1

x²/1 + y² / (1/4) = 1

The center (h, k) = (0, 0).

a² = 1
a = 1

b² = 1/4
b = 1/2

Answer 5

It is not meant to be solved. It is just a description of a set of points.

If you want to use the set to graph it. Pick an x and then solve for y^2. Use both the positive and negative values for y that you get. That will give you both halves of the ellipse.

Answer 6

Divide thru by 4 and take d square root of each side.