How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $x^2/9-y^2/25=1$ ?

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Answer 1 · 2017-12-26T02:12:23

vertices $(3,0) and (-3.0)$
Foci $(sqrt34,0) and (-sqrt34, 0)$
Aymptotes $y=5/3 x and y= - 5/3 x$

Given -

$x^2/9-y^2/25=1$

This hyperbola equation is in the form

$x^2/a^2-y^2/b^2=1$

If this is the case then

Its vertices are $(a, 0) and (-a,0)$
Its foci are $(c,0) and (-c,0)$

Its asymptotes are $y=b/a x and y = - b/a x$

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Then we have to find the values of $a,b and c$ from the given equation.

$a=sqrt(9)=3$
$b=sqrt25=5$
$c^2=a^2+b^2$
$c=+-sqrt (9+25)=+-sqrt34$

Then
vertices $(3,0) and (-3.0)$
Foci $(sqrt34,0) and (-sqrt34, 0)$
Aymptotes $y=5/3 x and y= - 5/3 x$

enter image source here

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola x^2/9-y^2/25=1x^2/9-y^2/25=1?