What are the foci of the ellipse with the equation 16x^2+4y^2=64? Graph the ellipse.

Please show all steps! I don't understand this. :/

Question

What are the foci of the ellipse with the equation 16x^2+4y^2=64? Graph the ellipse.

Please show all steps! I don't understand this. :/

anonymous · Answer

@hartnn

anonymous · Answer

@whpalmer4

anonymous · Answer

c^2=a^2-b^2=16-4=12

c=√12
Foci: (0, 0±c)=(0,0±√12)=(0,-√12) and (0,√12)

is this work correct, @Mertsj?

mertsj · Answer

You first need to divide both sides by 64 to get it into the right form.  Did you do that?

mertsj · Answer

$$\frac{x^2}{4}+\frac{y^2}{16}=1$$

anonymous · Answer

Yeah I did that. Is writing it in the form I have above still correct?

mertsj · Answer

So x^2 is 16 and b^2 is 4 and c is sqrt12 or 2sqrt3
However, the big number is under the y so the major axis is vertical and the foci are c units above and c units below the center.

mertsj · Answer

So if you just simplify the sqrt 12, you are done.  And you did a good job.

whpalmer4 · Answer

Indeed!

anonymous · Answer

Thank you both! :)

mertsj · Answer

yw