Step-by-step explanation:

This is the equation of the ellipse. Since the denominator is greater for the y values, we have a vertical ellipse. Remember a>b, so a
The formula for the foci of the vertical ellipse is
(h,k+c) and (h,k-c).
where c is
Our center (h,k) is (2, -5)

Here a^2 is 9, b^2 is 4.



So our foci is

and

Use pythagora's theorem to test if it is a right triangle
c^2 = a^2 + b^2
65^2 = 52^2 + 43^2
4225 = 2704 + 1849
4225 =/= 4553
therefore it is not a right triangle as it does not comply to pythagora's theorem
What is this question asking?
maybe try 3.87
Hello :
<span>x² + y² + 14x + 2y + 14 = 0
(x²+14x)+(y²+2y)+14=0
(x²+2(7)(x) +7²) -7² +( y²+(2)(1)y+1²)-1² +14 = 0
(x+7)² +(y+1)² = 6²
the center : (-7,-1) and </span><span>length of the radius is : 6</span>