How do you graph the ellipse #(x-2)^2/4+(y+3)^2/9=1# and find the center, the major and minor axis, vertices, foci and eccentricity?

1 Answer

Answer:

Explanation given below

Explanation:

The given equation of ellipse:

#\frac{(x-2)^2}{4}+\frac{(y+3)^2}{9}=1#

#\frac{(x-2)^2}{2^2}+\frac{(y+3)^2}{3^2}=1#

Above equation of ellipse is in standard form #(x-h)^2/a^2+(y-k)^2/b^2=1\quad (\forall \ a < b)#

The center of ellipse : # (h, k)\equiv (2, -3)#

Major axis: #2b=2(3)=6#

Minor axis: #2a=2(2)=4#

The vertices : #(h\pma, k)# & #(h, k\pmb)#

#(2\pm2, -3)# & #(2, -3\pm3)#

Eccentricity #e=\sqrt{1-a^2/b^2}=\sqrt{1-2^2/3^2}=\sqrt5/3#

Focii : #(h, k\pm be)#

#(2, -3\pm 3\cdot \sqrt5/3)\equiv(2, -3\pm\sqrt5)#

The equations of axes of ellipse: #y+3=0# & #x-2=0#

Plot the axes of ellipse. Specify the center #(2, -3)#, vertices & focii on the corresponding axes. Draw a free hand curve passing through the vertices & the points of intersection of ellipse with coordinate axes to get graph.