Question #129234
A hyperbola has the equation
16x²-9y²+64x-90y =305
Find its center,foci,vertices, length of latus rectum and equation of the asymptotes
1
Expert's answer
2020-08-20T18:10:43-0400

16x29y2+64x90y=30516(x2+4x)9(y2+10y)=30516(x2+4x+4)9(y2+10y+25)=305+6422516x^2-9y^2+64x-90y=305\\ 16(x^2+4x)-9(y^2+10y)=305\\ 16(x^2+4x+4)-9(y^2+10y+25)=305+64-225

16(x+2)29(y+5)2=144(x+2)29(y+5)216=116(x+2)^2-9(y+5)^2=144\\ \frac{(x+2)^2}{9}-\frac{(y+5)^2}{16}=1

h=-2 , k=5,

a=9=3b=16=4c=a2+b2=9+16=25=5a=\sqrt9=3\\ b=\sqrt{16}=4\\ c=\sqrt{a^2+b^2}=\sqrt{9+16}=\sqrt{25}=5


Centre: (-2,-5),

vertices: (-5,5) and (1,-5),

foci: (-7,-5) and (3,-5),

asymptotes:

y=5±4(x+2)3y=-5\pm4\frac{(x+2)}{3}


Length of latus rectrum =2b2a\frac{2b^2}{a}\\

=2163=323=10.66=\frac{2*16}{3}\\ =\frac{32}{3}\\ =10.66


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