How do you sketch #4x ^ { 2} + y ^ { 2} - 8x - 4y - 28= 0#?

1 Answer
Jul 16, 2017

Please see below.

Explanation:

#4x^2+y^2-8x-4y-28=0# is the equation of an ellipse, as can be seen from the following

#4x^2+y^2-8x-4y-28=0#

#hArr4(x^2-2x+1)-4+(y^2-4y+4)-4-28=0#

or #4(x-1)^2+(y-2)^2=36#

or #(x-1)^2/9+(y-2)^2/36=1# or #(x-1)^2/3^2+(y-2)^2/6^2=1#

an ellipse, whose center is #(1,2)# and major axis is #2xx6=12# parallel to #y#-axis and minor axis is #2xx3=6# parallel to #x#-axis.

End points of major axis are #(1,2+-6)# i.e. #(1,-4)# and #(1,8)#

End points of minor axis are #(1+-3,2)# i.e. #(-2,2)# and #(4,2)#

It appears as follows:

graph{(4x^2+y^2-8x-4y-28)((x-1)^2+(y-2)^2-0.06)((x-1)^2+(y+4)^2-0.06)((x-1)^2+(y-8)^2-0.06)((x+2)^2+(y-2)^2-0.06)((x-4)^2+(y-2)^2-0.06)=0 [-19.5, 20.5, -8.32, 11.68]}