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

Question

hero · Answer

pk, solve this

hero · Answer

lol

hero · Answer

He ran away

dumbcow · Answer

write it in standard form
$$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} =1$$
where vertices are center +- a

anonymous · Answer

dumbcow is correct, and @hero, I was helping coolaid earlier.

hero · Answer

You don't have to defend yourself pk

dumbcow · Answer

use completing the square
rearrange equation to group x's and y's
$$4x^{2}-80x +y^{2}-6y = -309$$
$$4(x^{2}-20x) +y^{2}-6y = -309$$
$$4(x^{2}-20x +100) +y^{2}-6y+9 = -309 +9 +4(100)$$
$$4(x-10)^{2} + (y-3)^{2} = 100$$
$$\frac{4(x-10)^{2}}{100} + \frac{(y-3)^{2}}{100} = 1$$
$$\frac{(x-10)^{2}}{5^{2}} + \frac{(y-3)^{2}}{10^{2}} = 1$$

dumbcow · Answer

a is always larger
a=10
b =5

dumbcow · Answer

center (10,3)
a is under y so major axis is vertical
vertices: (10,3+-10)