How do you find the coordinates of the center, foci, lengths of the major and minor axes given #y^2/18+x^2/9=1#?

Question

Elias Ackerman · Accepted Answer

The center is #=(0,0)#
The lengh of the major axis is #=6sqrt2#
The length of the minor axis is #=6# The general equation of the ellipse is #(y-h)^2/a^2+(x-k)^2/b^2=1# The center is #=(k,h)# The foci are #(k,h+-c)# The center is #=(0,0)# The lengh of the major axis is #=2*sqrt18=6sqrt2# The length of the minor axis is #=2sqrt9=6# To determine the foci, we need #c=sqrt(18-9)=3# Therefore, the foci are F#=(0,3)# and F'#(0,-3)# graph{(y^2/18)+(x^2/9)=1 [-11.25, 11.25, -5.625, 5.625]}

Greyson Brown · Answer

undefined Given the equation $ \frac{y^2}{18} + \frac{x^2}{9} = 1 $, this is the equation of an ellipse centered at the origin ($0,0$). The standard form of an ellipse centered at the origin is $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a $ is the length of the semi-major axis and $ b $ is the length of the semi-minor axis.

Comparing the given equation to the standard form, we find that $ a^2 = 9 $ and $ b^2 = 18 $.

1. The center of the ellipse is at the origin, $ (0,0) $.

2. The lengths of the major and minor axes are determined by $ 2a $ and $ 2b $, respectively. So, for this ellipse:
   - Length of major axis = $ 2a = 2\sqrt{9} = 6 $
   - Length of minor axis = $ 2b = 2\sqrt{18} = 6\sqrt{2} $

3. To find the foci, the distance from the center to each focus is given by $ c = \sqrt{a^2 - b^2} $:
   - $ c = \sqrt{9 - 18} = \sqrt{-9} = 3i $ (Imaginary number because the square root of a negative number)
   - The foci are at $ (0, 3i) $ and $ (0, -3i) $ in the complex plane.

So, the coordinates of the center are $ (0,0) $, the lengths of the major and minor axes are 6 and $ 6\sqrt{2} $ respectively, and the foci are at $ (0, 3i) $ and $ (0, -3i) $.