Question

Find an equation that models the path of a satelite if its path is a hyperbola, a=45,000 km, and c= 71,000 km. assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

Please show your work so i can understand!!

Answer 1

Answer:16000

Explanation:

b

Answer 2

so first off, we know the hyperbola has a horizontal traverse axis, so that means equation wise, the positive fraction is the one with the "x" variable on it, center is the origin, meaning (h,k) is just (0,0), and we know the value for the distances "a" and "c", let's find "b".

`\textit{hyperbola, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad √( a ^2 + b ^2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0\\ c=71000\\ a=45000 \end{cases}\implies \cfrac{(x- 0)^2}{ 45000^2}-\cfrac{(y- 0)^2}{ b^2}=1 \\\\[-0.35em] ~\dotfill`

`\stackrel{c}{71000}=\sqrt{\underset{ a }{45000^2}+b^2}\implies 71000^2=45000^2+b^2 \\\\\\ 71000^2-45000^2=b^2\implies √(71000^2-45000^2)=b\implies 54918.12\approx b \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(x- 0)^2}{ 45000^2}-\cfrac{(y- 0)^2}{ 54918.12^2}=1\implies {\Large \begin{array}{llll} \cfrac{x^2}{ 45000^2}-\cfrac{y^2}{ 54918.12^2}=1 \end{array}}`