I was given this ellipse 4x^2 + 9y^2 - 8x + 36y + 4 = 0.
Im really unsure how to put it in it's standar form but i have to i beleive to help me.
I actually need to find the center and foci of the elliple, how would i go about doing that?
I was given this ellipse 4x^2 + 9y^2 - 8x + 36y + 4 = 0.
Im really unsure how to put it in it's standar form but i have to i beleive to help me.
I actually need to find the center and foci of the elliple, how would i go about doing that?
Hello, 14041471!
You've never done one of these before . . . ever?
(a) Write in standard form: .$\displaystyle 4x^2 + 9y^2 - 8x + 36y + 4 \:=\: 0$
(b) Find the center and foci of the ellipse.
(a) We have: .$\displaystyle 4x^2 - 8x + 9y^2 + 36y \:=\:-4$
Factor: .$\displaystyle 4(x^2-2x \quad) + 9(y^2 - 4y \quad) \:=\:-4$
Complete the square: .$\displaystyle 4(x^2-2x \:{\color{blue}+ 1}) + 9(y^2 + 4y \:{\color{red}+\: 4}) \;=\;-4 \:{\color{blue}+\:4} \:{\color{red}+\: 36} $
Factor: .$\displaystyle 4(x-1)^2 + 9(y+2)^2 \:=\:36 $
Divide by 36: .$\displaystyle \frac{4(x-1)^2}{36} + \frac{9(y+2)^2}{36} \:=\:\frac{36}{36} \quad\Rightarrow\quad \boxed{\frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} \:=\:1}$
(b) The center is: .$\displaystyle \boxed{C(1,-2)}$
We have: .$\displaystyle a = 3,\:b = 2$
This is a "horizontal" ellipse; the foci on are the horizontal axis,
. . to the left and right of the center.
Since $\displaystyle c^2 \:=\:a^2-b^2$, we have: .$\displaystyle c^2 \:=\:9 - 4 \:=\:5\quad\Rightarrow\quad c \:=\:\pm\sqrt{5}$
The foci are: .$\displaystyle \boxed{F\left(1\pm\sqrt{5},\:-2\right)}$
Cheers. I understand that concept alot better now, I also need to find the length of the major and minor axis of the same ellipsis "4x^2 + 9y^2 - 8x + 36y + 4 = 0" how would i do that?
Edit: foget that i know now that
2a = 6 = big axis
and 2b = 4 = small axis