Write the standard form of the ellipse x^2+4y^2+6x-8y+9=0
I'm not sure how to start it, I think I need to set it equal to one because eventually it needs to end up equal to one but I'm not certain. I know the answer is (x+3)^2/4 + (y-1)^2/1=1 but have no idea how to get there. Any help?
Show steps please.
you have to complete the squares. Rearrange things a bit and you have
x^2+6x + 4y^2-8y = -9
x^2+6x + 4(y^2-2y) = -9
Now complete the squares, and be sure to make the same changes to both sides of the equation:
x^2+6x+9 + 4(y^2-2y+1) = -9+9+4
(x+3)^2 + 4(y-1)^2 = 4
(x+3)^2/4 + (y-1)^2 = 1
Thank you very much...
x^2 +y^2-8y-9=0
Well, let's clown around with this equation and see if we can figure it out!
To begin, we want to get the equation into a specific form, but before we do that, we need to rearrange a bit. So let's gather the x-terms and y-terms separately:
x^2 + 6x + 4y^2 - 8y + 9 = 0
Now, it's time for a clown trick! We want to complete the square for both the x-terms and y-terms. Let's focus on the x-terms first:
Take half of the coefficient of 'x' (which is 6) and square it: (6/2)^2 = 9.
Now, add and subtract this value inside the parentheses:
x^2 + 6x + 9 + 4y^2 - 8y + 9 = 0 - 9 + 9
Simplify that mess:
(x^2 + 6x + 9) + 4y^2 - 8y + 9 = 0
Hey, wait a minute! This looks like a perfect square!
(x + 3)^2 + 4y^2 - 8y + 9 = 0
Clownmath is magical, so let's work on completing the square for the y-terms now:
Take half of the coefficient of 'y' (which is -8) and square it: (-8/2)^2 = 16.
Now, add and subtract this value inside the parentheses:
(x + 3)^2 + 4y^2 - 8y + 16 - 16 + 9 = 0
Simplify that mess once again:
(x + 3)^2 + 4(y^2 - 2y + 4) - 7 = 0
But wait, there's more clownery happening here! The term inside the parentheses, y^2 - 2y + 4, is also a perfect square!
It's actually (y - 1)^2!
So let's use this to rewrite the equation:
(x + 3)^2 + 4(y - 1)^2 - 7 = 0
Almost there! Now, we want to isolate the squares by dividing all terms by 4:
(x + 3)^2/4 + (y - 1)^2/1 - 7/4 = 0
And there we have it, a beautifully structured clown form for the ellipse equation:
(x + 3)^2/4 + (y - 1)^2/1 = 7/4
See? Clowns can be helpful too!
To write the equation of the ellipse in standard form, you will need to complete the square for both the x and y terms.
1. Start by grouping the x terms and the y terms separately:
(x^2 + 6x) + 4(y^2 - 2y) + 9 = 0
2. To complete the square for the x terms, take half of the coefficient of x (which is 6), square it, and add it to both sides of the equation:
(x^2 + 6x + 9) + 4(y^2 - 2y) + 9 = 9
(x + 3)^2 + 4(y^2 - 2y) + 9 = 9
3. To complete the square for the y terms, take half of the coefficient of y (which is -2), square it, and add it to both sides of the equation:
(x + 3)^2 + 4(y^2 - 2y + 1) + 9 = 9 + 4
(x + 3)^2 + 4(y^2 - 2y + 1) + 9 = 13
4. Simplify the equation further:
(x + 3)^2 + 4(y - 1)^2 + 9 = 13
5. Subtract 9 from both sides of the equation:
(x + 3)^2 + 4(y - 1)^2 = 4
6. Divide both sides of the equation by 4 to isolate the term with the variable:
(x + 3)^2/4 + (y - 1)^2 = 1
Therefore, the standard form of the ellipse is (x + 3)^2/4 + (y - 1)^2/1 = 1.