# Thread: I believe this has to do with conics (not sure)

1. ## I believe this has to do with conics (not sure)

Hello,

I have been staring at this question for awhile now, and I just can't seem to figure out what to do for either of the questions. Help would be very much appreciated.

- Thanks

2. Originally Posted by yaphetkotto
Hello,

I have been staring at this question for awhile now, and I just can't seem to figure out what to do for either of the questions. Help would be very much appreciated.

- Thanks

Let $C(x_C, y_C)$ denote the center of the ellipse. Then the equation of the ellipse is:

$\dfrac{(x-x_C)^2}{a^2} +\dfrac{(y-y_C)^2}{b^2} =1$

The length of the semi-axis pointing in x-direction is 3 and the length of the semi-axis pointing in y-direction is $\frac92$

Plug in these values to get the equation of the ellipse:

$\dfrac{(x-3)^2}{3^2} +\dfrac{(y-\frac92)^2}{\left(\frac92 \right)^2} =1~~\implies~~\dfrac{(x-3)^2}{3^2} +\dfrac{4(y-\frac92)^2}{9^2} =1$

The second example has to be done in just the same way. Only the coordinates of the center has changed to C(-6, 0)

3. Thank you so much,

What was once foggy is now so clear.