Question

Write the standard form equation of the ellipse shown in the graph, and identify the foci.

Answer 1

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3)  and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

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

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6  and b = 4

The standard equation  becomes

$\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1$

$\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1$

Foci  are (h,k+c)  and (h,k-c)

$c=\sqrt{a^2-b^2}$

Plug in the a=6  and b=4

$c=\sqrt{6^2-4^2}$

 $c=\sqrt{20}$

  $c=2\sqrt{5}$, we know h=-2  and k=-3

Foci  are   $(-2,-3+2\sqrt{5})$  and  $(-2,-3-2\sqrt{5})$

Option A is correct