How do you put #9x^2+25y^2-54x-50y-119=0# in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

1 Answer
Sep 16, 2017

Answer:

Please see below.

Explanation:

#(x-h)^2/a^2+(y-k)^2/b^2=1# is the equation of an ellipse,

whose center is #(h,k)#, #2a# and #2b# are major axis and minor axis (if #a>b#, but if #a< b#, it is reverse). In the former case major axis is parallel to #x#-axis and in latter case major axis is parallel to minor axis.

vertices are #(h,k+-b)# and #(h+-a,k)#

Eccentricity #e# can be obtained using #e=sqrt(1-("minor axis")^2/("major axis")^2)#

and focii are at a distance of #ae# on either side of center along major axis (or #be# if major axis is parallel to #y#-axis).

We can write #9x^2+25y^2-54x-50y-119=0# as

#(9x^2-54x+81)+(25y^2-50y+25)=119+81+25=225#

or #(3x-9)^2/225+(5y-5)^2/225=1#

or #(x-3)^2/25+(y-1)^2/9=1# or #(x-3)^2/5^2+(y+1)^2/3^2=1#

center is #(3,1)#, major axis is #2xx5=10#, which is parallel to #x#-axis and minor axis is #2xx3=6#

Vertices are #(3,1+-3)# and #(3+-5,1)# i.e. #(3,4)#, #(3,-2)#, #(8,1)# and #(-2,1)#.

Eccentricity is #e=sqrt(1-(3/5)^2)=sqrt(9/25)=4/5# and #ae=4#

and focii are #(3+-4,1)# i.e. #(-1,1)#and #(7,1)#.

They appear as follows:

graph{(9x^2+25y^2-54x-50y-119)((x-3)^2+(y-1)^2-0.03)((x-3)^2+(y-4)^2-0.03)((x-3)^2+(y+2)^2-0.03)((x-8)^2+(y-1)^2-0.03)((x+2)^2+(y-1)^2-0.03)((x-7)^2+(y-1)^2-0.03)((x+1)^2+(y-1)^2-0.03)=0 [-7.14, 12.86, -3.78, 6.22]}