Question

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)?

Answer 1

`(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]}