Question

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ±(5/4)x

Answer 1

The transverse axis is parallel to y axis . (h,k) is the centre (say) - (0,0)

Answer 2

Now,

a = distance from the center to the vertices along the transverse axis
b = distance from the center to the endpoints of the conjugate axis
c = distance from the center to the foci along the transverse axis

it is given that, a= 10 here.

The relationship established between a, b and c is :

\(\bf 10^2 + b^2 = c^2\)

Given, slope of asympotes is \(\bf{\pm \cfrac{5}{4} }\)

a = is a multiple of 5
b = is a multiple of 4.

Answer 3

As, a = 10. So, by symmetry , b = 8

Answer 4

\(\bf{10^2 + 8^2 = c^2}\)
\(\bf{100+64=c^2}\)
\(\bf{164 = c^2}\)
\(\bf{c = 2\sqrt{41}}\)

Answer 5

Can you write equation of hyperbola now?