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=1x29−y225=1?

1 Answer
Nallasivam V
Dec 26, 2017

vertices (3,0) and (-3.0)(3,0)and(−3.0)
Foci (sqrt34,0) and (-sqrt34, 0)(√34,0)and(−√34,0)
Aymptotes y=5/3 x and y= - 5/3 xy=53xandy=−53x

Explanation:

Given -

x^2/9-y^2/25=1x29−y225=1

This hyperbola equation is in the form

x^2/a^2-y^2/b^2=1x2a2−y2b2=1

If this is the case then

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

Its asymptotes are y=b/a x and y = - b/a xy=baxandy=−bax

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

a=sqrt(9)=3a=√9=3
b=sqrt25=5b=√25=5
c^2=a^2+b^2c2=a2+b2
c=+-sqrt (9+25)=+-sqrt34c=±√9+25=±√34

Then
vertices (3,0) and (-3.0)(3,0)and(−3.0)
Foci (sqrt34,0) and (-sqrt34, 0)(√34,0)and(−√34,0)
Aymptotes y=5/3 x and y= - 5/3 xy=53xandy=−53x

enter image source here

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