Math  /  Geometry

Question4cr=(0,2)(0,2)c2=164c2=±12c=±234 \left\lvert\, \begin{array}{l} c-r=(0,2)(0,-2) \\ c^{2}=16-4 \\ c^{2}= \pm \sqrt{12} \\ c= \pm 2 \sqrt{3} \end{array}\right. 3) (x1)29+(y+5)24=1\frac{(x-1)^{2}}{9}+\frac{(y+5)^{2}}{4}=1 c=(1,5)c=(1,-5) 5) x2+9y2+6x90y+225=0x^{2}+9 y^{2}+6 x-90 y+225=0

Studdy Solution

STEP 1

What is this asking? We need to find the center of the ellipse x2+9y2+6x90y+225=0x^2 + 9y^2 + 6x - 90y + 225 = 0. Watch out! Don't forget to complete the square correctly for both xx and yy terms!

STEP 2

1. Rewrite the equation
2. Complete the square for *x*
3. Complete the square for *y*
4. Rewrite in standard form

STEP 3

Let's **group** the xx terms, the yy terms, and move the constant to the other side.
This sets us up perfectly to complete the square! x2+6x+9y290y=225x^2 + 6x + 9y^2 - 90y = -225

STEP 4

**Factor out** the 9\textbf{9} from the yy terms: x2+6x+9(y210y)=225x^2 + 6x + 9(y^2 - 10y) = -225 This makes completing the square for the yy terms much cleaner!

STEP 5

To complete the square for the xx terms, we take **half** of the coefficient of the xx term, which is 66, so half is 3\textbf{3}, and **square it**: 32=93^2 = \textbf{9}.
We **add** this to both sides: x2+6x+9+9(y210y)=225+9x^2 + 6x + 9 + 9(y^2 - 10y) = -225 + 9 (x+3)2+9(y210y)=216(x+3)^2 + 9(y^2 - 10y) = -216Now the xx terms are a perfect square!

STEP 6

Now, for the yy terms inside the parentheses!
Take **half** of the coefficient of the yy term, which is 10-10, giving us -5\textbf{-5}. **Squaring** that gives us (5)2=25(-5)^2 = \textbf{25}.
We **add** this inside the parentheses.
Remember, since it's multiplied by 99, we're really adding 925=2259 \cdot 25 = \textbf{225} to the left side.
So, we must **add** 225\textbf{225} to the right side as well! (x+3)2+9(y210y+25)=216+225(x+3)^2 + 9(y^2 - 10y + 25) = -216 + 225 (x+3)2+9(y5)2=9(x+3)^2 + 9(y-5)^2 = 9Woohoo! Both xx and yy terms are perfect squares now!

STEP 7

To get the standard form of an ellipse equation, we need the right side to be 1\textbf{1}. **Divide** both sides by 9\textbf{9}: (x+3)29+9(y5)29=99\frac{(x+3)^2}{9} + \frac{9(y-5)^2}{9} = \frac{9}{9} (x+3)29+(y5)21=1\frac{(x+3)^2}{9} + \frac{(y-5)^2}{1} = 1

STEP 8

This ellipse is centered at (3,5)(-3, 5).
Remember, the standard form is (xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, where (h,k)(h, k) is the center.

STEP 9

The center of the ellipse is (3,5)(-3, 5).

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