Find center, focii and intercepts on #x#-axis and #y#-axis of ellipse #16x^2+9y^2+64x-18y-71=0#?

1 Answer
Jul 7, 2017

Answer:

Center is #(-2,1)#, focii are #(-2,1-sqrt7)# and #(-2,1+sqrt7)#. Intercepts on #x#-axis are #1+-(4sqrt5)/3# and intercepts on #y#-axis are #-2+-(3sqrt15)/4#.

Explanation:

The general form of equation of ellipse is

#(x-h)^2/a^2+(y-k)^2/b^2=1#, whose center is #(h,k)# and major axis is along #x#-axis if #a>b# and along #y#-axis if #a< b#.

Further eccentricity #e=sqrt(1-b^2/a^2)#, if #a>b# or #e=sqrt(1-a^2/b^2)#, if #a< b#.

Focii are at major axis at a distance of #+-ae# or #=+-be#, again depending on #a>b# or #b> a#.

We can express #16x^2+9y^2+64x-18y-71=0# as

#16x^2+64x+9y^2-18y-71=0#

or #16(x^2+4x+4)+9(y^2-2y+1)-64-9-71=0#

or #16(x+2)^2+9(y-1)^2=144#

or #(x+2)^2/3^2+(y-1)^2/4^2=1#

Hence center is #(-2,1)# and major axis along #y#-axis is #8# and minor axis along #x#-axis is #6#.

Eccentricity is #e=sqrt(1-(3/4)^2)=sqrt7/4#

and as #be=sqrt7#, focii are #(-2,1+-sqrt7)#

Intercepts on #x#-axis can be found by putting #y=0# i.e. #9y^2-18y-71=0# #y=(18+-sqrt(324-4*9*(-71)))/18=1+-sqrt(324+2556)/18=1+-24sqrt5/18=1+-(4sqrt5)/3#

Intercepts on #y#-axis can be found by putting #x=0# i.e. #16x^2+64x-71=0# #x=(-64+-sqrt(64^2-4*16*(-71)))/32=-2+-sqrt(4096+4544)/32=-2+-24sqrt15/32=-2+-(3sqrt15)/4#

graph{(16x^2+9y^2+64x-18y-71)((x+2)^2+(y-1-sqrt7)^2-0.03)((x+2)^2+(y-1+sqrt7)^2-0.03)=0 [-12.21, 7.79, -4.36, 5.64]}