Ellipse to Standard Form
This is the equation to the ellipse, $9x^2+4y^2-72x+40y+208 = 0$, and I need it in standard form. I can't figure this one out.
Could this be the answer? $\frac{(x-4)^2}{9} + \frac{(y+5)^2}{4} = 1$
The other options are:
$\frac{(x-4)^2}{4} + \frac{(y-5)^2}{9} = 1$
$\frac{(x-4)^2}{4} + \frac{(y+5)^2}{9} = 1$
$(x-4)^2 + (y+5)^2 = 36$
$\endgroup$ 33 Answers
$\begingroup$its 2nd option
$\frac{(x-4)^2}{4} + \frac{(y+5)^2}{9} = 1$
$\endgroup$ $\begingroup$The following is correct.
$$\frac{(x-4)^2}{4} + \frac{(y+5)^2}{9}= 1$$
$\endgroup$ $\begingroup$To convert a "general form" equation to "standard form", it will help to collect the "x-" and "y-terms", then factor out the quadratic coefficients before "completing the squares":
$$9x^2 \ -\ 72x \ + \ 4y^2 \ + \ 40y \ = \ -208 \ \ \Rightarrow \ \ 9 (x^2 \ -\ 8x) \ + \ 4( y^2 \ + \ 10y) \ = \ -208$$
$$\Rightarrow \ \ 9 (x^2 \ -\ 8x \ + \ 16) \ + \ 4( y^2 \ + \ 10y \ + 25 ) \ = \ -208 \ + \ 9 \cdot 16 \ + \ 4 \cdot 25 \ . $$
From there, you can set up the binomial squares in the "standard form" and divide through by the number on the right-hand side to find your result.
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