We want to identify the center, vertices, and foci of the ellipse. To do so, we will rewrite the given equation just a little bit.

\frac{( x + 7 ) ^{2}}{2 2 5} + \frac{( y + 1 ) ^{2}}{1 4 4} = 1

Write as a power

\frac{( x + 7 ) ^{2}}{1 5 ^{2}} + \frac{( y + 1 ) ^{2}}{1 2 ^{2}} = 1

$a = - (- a)$

\frac{( x - ( - 7 ) ) ^{2}}{1 5 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{1 2 ^{2}} = 1

Notice that the denominator of the expression containing the

x -

variable is greater than the denominator of the expression that contains the

y -

variable. Therefore, we have a horizontal ellipse. Let's recall the main characteristics of this type of ellipse.

Let's consider our equation one more time.

\frac{( x - ( - 7 ) ) ^{2}}{1 5 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{1 2 ^{2}} = 1

We can see that

a = 15,

b = 12,

h = - 7,

and

k = - 1 .

The only value we do not know is

c .

Let's find it!

c^{2} = a^{2} - b^{2}

$b = 12$ , $a = 15$

c^{2} = 15^{2} - 12^{2}

▼

Solve for

c

Calculate power

c^{2} = 225 - 144

Subtract term

c^{2} = 81

$LHS = RHS$

c = 9

Note that we only took the principal root, because to find the foci we will add and subtract the value of

c .

Therefore, its sign is irrelevant. Finally, we can write the desired information.

Center	Vertices	Foci
$(- 7, - 1)$	$(- 7 \pm 15, - 1)$ $⇓$ $(8, - 1) and (- 22, - 1)$	$(- 7 \pm 9, - 1)$ $⇓$ $(2, - 1) and (- 16, - 1)$

Exercise