# 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
$\displaystyle 2x^2 + 3y^2 - 4x - 12y +13 = 0$

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
$\displaystyle 2x^2 + 3y^2 - 4x - 12y +13 = 0$

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

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

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

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

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

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

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

Now shifting the center to (1,2) weget

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

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. $\displaystyle 2(x^2 - 2x) + 3(y^2 - 4y) = -13$
$\displaystyle 2(x^2 - 2x) + 3(y^2 - 4y) = \frac{-13}{2} . 2$
$\displaystyle 4(x^2 - 2x) + 6(y^2 - 4y) = -36$
$\displaystyle (x^2 - 2x) + \frac{3}{2}(y^2 - 4y) = -9$
$\displaystyle \frac{1}{3}(x^2 - 2x) + \frac{1}{6}(y^2 - 4y) = -3$

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

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

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

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