Question

What are the vertices of the ellipse given by the equation `4x^2 + y^2 – 80x – 6y + 309 = 0`?

Answer 1

In this way, since the equation of the ellipse is this:

`(x-x_c)^2/a^2+(y-y_c)^2/b^2=1`,

`4x^2+y^2–80x–6y+309=0rArr`

`4(x^2-20x)+(y^2-6y)+309=0rArr`

`4(x^2-20x+100-100)+(y^2-6y+9-9)+309=0rArr`

`4(x^2-20x+100)-400+(y^2-6y+9)-9+309=0rArr`

`4(x-10)^2+(y-3)^2=100`

`(x-10)^2/25+(y-3)^2/100=1rArr`

So the center is in `C(10,3)` and the semi-axes are `a=5` and `b=10`.

The vertices are from the center horizontally right of `a`, horizontally left of `a`, verticalli up of `b`, vertically down of `b`.

So:

`V_1(15,3)`

`V_2(5,3)`

`V_3(10,13)`

`V_4(10,-7)`.

The graph is:

graph{(x-10)^2/25+(y-3)^2/100=1 [-15.96, 29.67, -8.02, 14.78]}