1. ## Equation Questions

1.If I have the equation (x+4)squared + (y+4)squared = 9

The center of the circle would be (-4,-4) correct?

2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)

The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?

Is there an easier way to write this equation?

Thanks.

2. Okay, I figured out the first part. Do any of you know number two.

3. Originally Posted by milosh
1.If I have the equation (x+4)squared + (y+4)squared = 9

The center of the circle would be (-4,-4) correct?

correct

2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)

The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?

Is there an easier way to write this equation?

yes, there is ...

$\textcolor{red}{4(x-20)^2 + 16(y-10)^2 = 1}$
...

4. Originally Posted by milosh
1.If I have the equation (x+4)squared + (y+4)squared = 9

The center of the circle would be (-4,-4) correct?

2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)
Do you mean "over 1/4" or "over 4"? If you actually mean "over 1/4" and "over 1/16", the equation is
$\frac{(x-20)^2}{1/4}+ \frac{(y-10)^2}{1/16}= 4(x-20)^2+ 16(y-10)^2= 1$.

If you meant "over 4" and "over 16", the equation is
$\frac{(x-20)^2}{4}+ \frac{(y-10)^2}{16}= 1$
which can also be written as
$16(x- 20)^2+ 4(y-10)^2= 64$
after multiplying both sides by 4(16)= 64.

The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?
That sentence doesn't make sense. If you mean "1/4 is the length of the semi-axis parallel to the x-axis and 1/16 is the length of the semi-axis parallel to the y-axis" and you really are talking about
$\frac{(x-20)^2}{1/4}+ \frac{(y-10)^2}{1/16}= 4(x-20)^2+ 16(y-10)^2= 1$
then the answer is "No". The lengths of the x and y semi-axes are $1/\sqrt{4}= 1/2$ and $1/\sqrt{16}= 1/4$ respectively. If you actually meant
$16(x- 20)^2+ 4(y-10)^2= 64$
then the answer is stilll "No"- the lengths of the x and y semi axes are $\sqrt{4}= 2$ and $\sqrt{16}= 4$, respectively,

Is there an easier way to write this equation?

Thanks.