Put the equation in standard form for ellipse

4x^2 +16x + 25y^2 + 250y + 541 = 0

thank you!

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- Jun 9th 2013, 10:16 AMMc3Ellipse Standard Form
Put the equation in standard form for ellipse

4x^2 +16x + 25y^2 + 250y + 541 = 0

thank you! - Jun 9th 2013, 03:27 PMShadowKnight8702Re: Ellipse Standard Form
The key to this is to factor out in such a way in order to form the standard equation of an ellipse. To do this, we complete the square. With the 4x^2 + 4x, you know that you need to add 16 in order to make it a square, so take out the 16 from the 541. You will get 4(x + 2)^2 + 25y^2 + 250y + 525 = 0. Do the same thing with the y, and you get 4(x + 2)^2 + 25(y + 5)^2 = 100. Divide both sides by 100 and you will get an ellipse. Just remember to complete the square with both x and y, and you will be fine. THis will in the end make an ellipse.

- Jun 9th 2013, 10:02 PMibduttRe: Ellipse Standard Form
I would recommend we follow following algorithm:

Step 1: Transfer the constant term to the RHS.

4x^2 +16x + 25y^2 + 250y = - 541

Step 2: Arrange terms on LHS such that we have terms containing x and y are separated

{4x^2 +16x }+ {25y^2 + 250y } = - 541

Step 3: Add constants to make the quadratic in x as well as y a perfect square, the same constant should be added on the RHS

{(2x)^2 +2*2*4 *x + (4)^2}+ {(5y)^2 + 2*5* 25* y + ( 25)^2 } = - 541+ (4)2 + ( 25)^2

(2x + 4)^2 + (5y+ 25)^2 = - 541+ 16 + 625 = 100

Step 4

Divide both sides by constant on RHS so that we have 1 on RHS

[(2x + 4)^2]/ 100 + [(5y+ 25)^2]/100 = 1

[4(x + 2)^2]/ 100 + [25(y+ 5)^2]/100 = 1

[(x + 2)^2]/ 25 + [(y+ 5)^2]/4 = 1

That is the equation of the ellipse