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

##### 1 Answer
May 1, 2015

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

${\left(x - {x}_{c}\right)}^{2} / {a}^{2} + {\left(y - {y}_{c}\right)}^{2} / {b}^{2} = 1$,

4x^2+y^2–80x–6y+309=0rArr

$4 \left({x}^{2} - 20 x\right) + \left({y}^{2} - 6 y\right) + 309 = 0 \Rightarrow$

$4 \left({x}^{2} - 20 x + 100 - 100\right) + \left({y}^{2} - 6 y + 9 - 9\right) + 309 = 0 \Rightarrow$

$4 \left({x}^{2} - 20 x + 100\right) - 400 + \left({y}^{2} - 6 y + 9\right) - 9 + 309 = 0 \Rightarrow$

$4 {\left(x - 10\right)}^{2} + {\left(y - 3\right)}^{2} = 100$

${\left(x - 10\right)}^{2} / 25 + {\left(y - 3\right)}^{2} / 100 = 1 \Rightarrow$

So the center is in $C \left(10 , 3\right)$ 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:

${V}_{1} \left(15 , 3\right)$

${V}_{2} \left(5 , 3\right)$

${V}_{3} \left(10 , 13\right)$

${V}_{4} \left(10 , - 7\right)$.

The graph is:

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