Answer
A) $c^2=49$, $c=7$
B) $c^2=0$, $c=0$
C) $c^2 =9$, $c=3$
D) $c^2=64$, $c=8$
E) $c^2=1$, $c=1$
F) $c^2=0$, $c=0$
G) $c^2=81$, $c=9$
H) $c^2=4$, $c=2$
Work Step by Step
The formula for an ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$, and the formula for a hyperbola is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} =1$.
Equations B and F have the same denominators for $x^2$ and $y^2$, so those are circles.
Equations C, E, and H have different denominators for $x^2$ and $y^2$, so those are ellipses.
Equations A, D, and G are looking for differences, so the formulas are for hyperbolas.
Circles and ellipses have $c^2=abs (a^2-b^2)$, and hyperbolas have $c^2=a^2+b^2$.
A)
$a^2=36$, $b^2=13$
$c^2=36+13$
$c^2=49$
$\sqrt {c^2} =\sqrt {49}$
$c=7$
B)
$a^2=4$, $b^2=4$
$c^2=abs(4-4)$
$c^2= abs 0$
$c^2=0$
$\sqrt {c^2} = \sqrt 0$
$c=0$
C)
$a^2=25$, $b^2=16$
$c^2=abs (a^2-b^2)$
$c^2=abs (25-16)$
$c^2=abs 9$
$c^2=9$
$\sqrt {c^2} = sqrt 9$
$c =3$
D)
$a^2=25$, $b^2=39$
$c^2=25+39$
$c^2=64$
$\sqrt {c^2} =\sqrt {64}$
$c=8$
E)
$a^2=17$, $b^2=18$
$c^2=abs (17-18)$
$c^2=abs (-1)$
$c^2=1$
$\sqrt {c^2} =sqrt 1$
$c =1$
F)
$a^2=36$, $b^2=36$
$c^2=abs(36-36)$
$c^2=abs 0$
$c^2=0$
$\sqrt {c^2} =\sqrt {0}$
$c = 0$
G)
$a^2=16$, $b^2=65$
$c^2=16+65$
$c^2=81$
$\sqrt {c^2} =\sqrt {81}$
$c=9$
H)
$a^2=144$, $b^2=140$
$c^2=abs (a^2-b^2)$
$c^2=abs (144-140)$
$c^2=abs (4)$
$c^2=4$
$\sqrt {c^2} =\sqrt 4$
$c=2$
