How do you find equation of ellipse with two vertices V1(7,12) and V2(7, -8), and passing through the point P(1,8)?

1 Answer
May 30, 2017

Answer:

#(4(x-7)^2)/225+(y-2)^2/100=1#

Explanation:

The equation of an ellipse with centre #(h,k)# and axis parallel to #x#-axis #2a# and axis parallel to #y#-axis #2b# is of the form #(x-h)^2/a^2+(y-k)^2/b^2=1#.

Here two vertices are #(7,12)# and #(7,-8)#, hence centre is #(7,(12-8)/2)# or #(7,2)# and as distance between vertices (with common abscissa) is #12-(-8)=20# and hence #2b=20# or #b=10#.

Hence equation is #(x-7)^2/a^2+(y-2)^2/100=1#

As it passes through #P(1,8)#

#(1-7)^2/a^2+(8-2)^2/100=1#

or #36/a^2+36/100=1#

or #36/a^2=1-36/100=64/100#

Hence #a^2=(36xx100)/64#

and #a=60/8=7.5# and equation of ellipse is

#(4(x-7)^2)/225+(y-2)^2/100=1#

Follwing is the graph with three given points marked.

graph{((4(x-7)^2)/225+(y-2)^2/100-1)((x-1)^2+(y-8)^2-0.04)((x-7)^2+(y-12)^2-0.04)((x-7)^2+(y+8)^2-0.04)=0 [-15, 35, -10, 15]}