Question

Please Help!!!

Answer 1

Your ellipse is given in the form:
$\frac{(x - xc)^{2} }{ a^{2} } + \frac{(y - yc)^{2} }{ b^{2} } = 1$

Indeed, you have:

$\frac{(x-3)^{2} }{64} + \frac{(y+5)^{2} }{100} = 1$

Therefore you can easily find:

a) the coordinates of the center:
in your case xc = 3 and yc = -5
hence C(3 , -5)

b) lenght of the major and minor axis:
in your case a² < b²:
a² = 64  ⇒ a = 8 (ATTENTION! This is the semi-minor axis)
b² = 100 ⇒ b = 10 (again, this is the semi-major axis)
Therefore,
minor axis 2a = 16 and major axis 2b = 20

c) coordinates of the foci:
Since you have a² < b², foci have the generic coordinates F₁₂ (xc , yc+/-c)
where c = √(b² - a²)

Let's compute c = √(100 - 64) = √36 = 6
yf₁ = -5 - 6 = -11
yf₂ = -5 + 6 = -1
Therefore:
F₁ (3, -11) and F₂(3, +1)

d) the graph is in the picture attached