the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
Answer:
G. <3 and <5 hope this helps
So I think the rule is that you divide two exponents with the same base, you keep the base and subtract the powers.
Answer:
The number of pages in each book varied more last summer than this summer
Step-by-step explanation:
