Answer
$\dfrac{x^2}{9}+\dfrac{y^2}{8}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(0,0)$
Focus: $(-1,0)$
Vertex: $(3,0)$
Because the $y$-coordinates of the vertex and focus are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k$ using the center:
$(h,k)=(0,0$
$h=0$
$k=0$
Determine $a$ using the vertex:
$(h+a,k)=(3,0)$
$(0+a,0)=(3,0)$
$(a,0)=(3,0)$
$a=3$
Determine $c$ using the focus:
$(h-c,k)=(-1,0)$
$(0-c,0)=(-1,0)$
$(-c,0)=(-1,0)$
$c=1$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=3^2-1^2$
$b^2=8$
$b=\sqrt 8=2\sqrt 2$
The equation of the ellipse is:
$\dfrac{x^2}{3^2}+\dfrac{y^2}{(2\sqrt 2)^2}=1$
$\dfrac{x^2}{9}+\dfrac{y^2}{8}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
