Find the vertices and foci of the hyperbola with equation x^2/4 - y^2/60 = 1

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 equationh=0, k=0so, Center of hyperbola is (0,0)Vertices of HyperbolaVertices of hyperbola can be found as:The first vertex can be found by adding h to aa^2 - 4 => a=2, h=0 and k=0So, first vertex is (h+a,k) = (2,0)The second vertex can be found by subtracting a from hso, second vertex is ( h-a,k) = (-2,0)Foci of HyperbolaFoci of hyperbola can be found asThe first focus of hyperbola can be found by adding c to hFinding 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=0The first focus is (h+c,k) = (8,0)The second focus is (h-c,k) = (-8,0)