Answer:
Vertices of hyperbola: (2,0) and (-2,0)
Foci of hyperbola: (8,0) and (-8,0)
Step-by-step explanation:
The given equation is:
[tex]\frac{x^2}{4}-\frac{y^2}{60}=1[/tex]
The standard form of equation of hyperbola is:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
Center of hyperbola is (h,k)
Comparing given equation with standard equation
h=0, k=0
so, Center of hyperbola is (0,0)
Vertices of Hyperbola
Vertices of hyperbola can be found as:
The first vertex can be found by adding h to a
a^2 - 4 => a=2, h=0 and k=0
So, first vertex is (h+a,k) = (2,0)
The second vertex can be found by subtracting a from h
so, second vertex is ( h-a,k) = (-2,0)
Foci of Hyperbola
Foci of hyperbola can be found as
The first focus of hyperbola can be found by adding c to h
Finding c (distance from center to focus):
[tex]c=\sqrt{a^2+b^2} \\c=\sqrt{(2)^2+(2\sqrt{15})^2}\\c=8[/tex]
So, c=8 , h=0 and k=0
The first focus is (h+c,k) = (8,0)
The second focus is (h-c,k) = (-8,0)
