Question

Given the ellipse (x−3)^2/4+(y−4)^2/25=1,Find the center point: List the four vertices

Answer 1

The equation of the ellipse of center (h, k) is

$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$

Its 4 vertices are

$(h,k+a)(h,k-a),(h+b,k),(h-b,k)$

Since the given equation is

$\frac{(x-3)^2}{4}+\frac{(y-4)^2}{25}=1$

Then by comparing it with the form above

$h=3,k=4$$b^2=4,b=2,-2$$a^2=25,a=5,-5$

Then we can find the 4 vertices using the rule above

$(h,k+a)=(3,4+5)=(3,9)$$(h,k-a)=(3,4-5)=(3,-1)$$\begin{gathered} (h+b,k)=(3+2,4)=(5,4) \\ (h-b,k)=(3-2,4)=(1,4) \end{gathered}$

The 4 vertices are

(3, 9), (3, -1), (5, 4), (1, 4)

The center is (3, 4)