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Question 610057: What are the directrices of the hyperbola given by the equation (y^2/49)-(x^2/9)=1?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What are the directrices of the hyperbola given by the equation
(y^2/49)-(x^2/9)=1?
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This is an equation of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2=1, (h,k)=(x,y) coordinates of center.
For given equation: (y^2/49)-(x^2/9)=1
center: (0,0)
a^2=49
a=√49=7
b^2=9
c^2=a^2+b^2=49+9=58
c=√58≈7.6
..
eccentricity, e=c/a=√58/7
directrices=±a/e=±a^2/c=±49/√58≈±6.43
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