problems with ellipse

**14041471** · January 8th 2009, 09:07 AM

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?

**Soroban** · January 8th 2009, 10:28 AM

Hello, 14041471!

You've never done one of these before . . . ever?

(a) Write in standard form: . $4x^2 + 9y^2 - 8x + 36y + 4 \:=\: 0$

(b) Find the center and foci of the ellipse.

(a) We have: . $4x^2 - 8x + 9y^2 + 36y \:=\:-4$

Factor: . $4(x^2-2x \quad) + 9(y^2 - 4y \quad) \:=\:-4$

Complete the square: . $4(x^2-2x \:{\color{blue}+ 1}) + 9(y^2 + 4y \:{\color{red}+\: 4}) \;=\;-4 \:{\color{blue}+\:4} \:{\color{red}+\: 36}$

Factor: . $4(x-1)^2 + 9(y+2)^2 \:=\:36$

Divide by 36: . $\frac{4(x-1)^2}{36} + \frac{9(y+2)^2}{36} \:=\:\frac{36}{36} \quad\Rightarrow\quad \boxed{\frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} \:=\:1}$

(b) The center is: . $\boxed{C(1,-2)}$

We have: . $a = 3,\:b = 2$
This is a "horizontal" ellipse; the foci on are the horizontal axis,
. . to the left and right of the center.

Since $c^2 \:=\:a^2-b^2$ , we have: . $c^2 \:=\:9 - 4 \:=\:5\quad\Rightarrow\quad c \:=\:\pm\sqrt{5}$

The foci are: . $\boxed{F\left(1\pm\sqrt{5},\:-2\right)}$

**14041471** · January 8th 2009, 11:08 AM

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

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