# Thread: Problem representing a 2 degree equation in the form of a ellipse

1. ## Problem representing a 2 degree equation in the form of a ellipse

I am having problem with representing a 2 degree eq in the form of a ellipse
$2x^2 + 3y^2 - 4x - 12y +13 = 0$

please provide me with the starting few steps or clues on how to solve this .I have already done the tranformation procedure and the result is this equation .I am stuck at this step as i dont know what to do next ?

2. Try completing the square for both x and y.

3. ## Please can u elaborate

Originally Posted by MathTooHard
Try completing the square for both x and y.

I didnt understand ur suggestion , could u please elaborate it

4. Originally Posted by zorro
I am having problem with representing a 2 degree eq in the form of a ellipse
$2x^2 + 3y^2 - 4x - 12y +13 = 0$

please provide me with the starting few steps or clues on how to solve this .I have already done the tranformation procedure and the result is this equation .I am stuck at this step as i dont know what to do next ?
Originally Posted by MathTooHard
Try completing the square for both x and y.
Originally Posted by zorro
I didnt understand ur suggestion , could u please elaborate it
If you rearrange the terms, we get

$2x^2-4x + 3y^2- 12y = -13\implies 2(x^2-2x)+3(y^2-4y)=-13$.

Now complete the square on x and y, just like MathTooHard suggested.

Does this clarify things?

5. ## Thanks for ur reply but u still not answered the question

Originally Posted by Chris L T521
If you rearrange the terms, we get

$2x^2-4x + 3y^2- 12y = -13\implies 2(x^2-2x)+3(y^2-4y)=-13$.

Now complete the square on x and y, just like MathTooHard suggested.

Does this clarify things?

Thanks for ur reply but u still not answered the question, what do u mean by square x and y
please can u provide me with the first few step ...I am not a so good with understanding the math verbiage

6. ## Is this correct

Originally Posted by MathTooHard
Try completing the square for both x and y.

From the clue given by u i have formulated a solution but i dont know if its correct or no

$2(x^2 - 2x) + 3(y^2 - 4y) = -13$
$2(x^2 - 2x + 1 - 1) + 3(y^2 - 4y + 4 - 4) = -13$
$2[(x^2 - 2x + 1) - 1] + 3[(y^2 - 4y + 4) - 4] = -13$
$2[(x^2 - 1)^2 - 1] + 3[(y^2 - 2)^2 - 4] = -13$
$2(x^2 - 1)^2 - 2 + 3(y^2 - 2)^2 - 12 = -13$
$2(x^2 - 1)^2 + 3(y^2 - 2)^2 - 14 = -13$
$2(x^2 - 1)^2 + 3(y^2 - 2)^2 = 14 -13$
$2(x^2 - 1)^2 + 3(y^2 - 2)^2 = 1$

$\frac{(x^2 - 1)^2}{\frac{1}{2}} + \frac{(y^2 - 2)^2}{\frac{1}{3}} = 1$

Now shifting the center to (1,2) weget

$\frac{X^2}{\frac{1}{2}} + \frac{Y^2}{\frac{1}{3}} = 1$

7. To learn how to complete the square to find the ellipse equation, try here.

8. ## Is the answer correct ?

Originally Posted by stapel
To learn how to complete the square to find the ellipse equation, try here.
$2(x^2 - 2x) + 3(y^2 - 4y) = -13$
$2(x^2 - 2x) + 3(y^2 - 4y) = \frac{-13}{2} . 2$
$4(x^2 - 2x) + 6(y^2 - 4y) = -36$
$(x^2 - 2x) + \frac{3}{2}(y^2 - 4y) = -9$
$\frac{1}{3}(x^2 - 2x) + \frac{1}{6}(y^2 - 4y) = -3$

$\frac{1}{3}(x^2 - 2x + 4) + \frac{1}{6}(y^2 - 4y + 16) = -3 \frac{1}{3}(4) + \frac{1}{6}(16)$

$\frac{1}{3}(x - 2)^2 + \frac{1}{6}(y - 4) = \frac{-9 + 4 + 8}{3}$

$\frac{(x - 2)^2}{3} + \frac{(y - 4)^2 }{6}= 1$