Finding standard equation of an Ellipse

Hello Everyone,

I can't figure out very well how to find the standard equation of the following ellipse.

$\displaystyle 25x^2+49y^2-300x+49y-1251/4=0$

I was trying the following steps.

$\displaystyle [25x^2-300x +- ?]+ [ 49y^2+49y +- ? ]= 1251/4$ Find squares (I'm not good at, finding squares)

Re: Finding standard equation of an Ellipse

Looking at the x part. If you want to get it into the form (ax+b)^{2} expanding it out might make it clearer. (ax+b)^{2}=a^{2}x^{2}+2abx+b^{2}

since the x^{2} part is 25x^{2} you know that a^{2}=25, a=5

The x part is -300. You know that 2ab=-300. and a=5 so b=-30

Lastly you need to balance the square by subtracting a constant.

$\displaystyle 25x^2-300x=(5x-30)^2+c$

For this to be true c must be equal to -900

Re: Finding standard equation of an Ellipse

Hello, vaironxxrd!

Before completing-the-square,

factor out the leading coefficients of the quadratic expressions.

Quote:

Find the standard equation of the following ellipse:

. . $\displaystyle 25x^2+49y^2-300x+49y-\tfrac{1251}{4}\:=\:0$

We have: . . $\displaystyle 25x^2 - 300x \quad+ 49y^2 + 49y\quad \:=\:\tfrac{1251}{4}$

Factor:.$\displaystyle 25(x^2-12x \qquad) + 49(y^2 + y \qquad) \:=\:\tfrac{1251}{4}$

. . . . . . $\displaystyle 25(x^2-12x + {\color{red}36}) + 49(y^2 + y +{\color{blue}\tfrac{1}{4}}) \:=\:\tfrac{1251}{4} + {\color{red}900} + {\color{blue}\tfrac{49}{4}}$

. . . . . . . . . . . . . . $\displaystyle 25(x-6)^2 + 49(y+\tfrac{1}{2})^2 \:=\:1225$

Divide by 1225:. . . . . $\displaystyle \frac{(x-6)^2}{49} + \frac{(y+\frac{1}{2})^2}{25} \;=\;1$