Answer to Question #129234 in Analytic Geometry for Sym

Question

A hyperbola has the equation
16x²-9y²+64x-90y =305
Find its center,foci,vertices, length of latus rectum and equation of the asymptotes

Answer 1

$16x^2-9y^2+64x-90y=305\\ 16(x^2+4x)-9(y^2+10y)=305\\ 16(x^2+4x+4)-9(y^2+10y+25)=305+64-225$

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

h=-2 , k=5,

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

Centre: (-2,-5),

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

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

asymptotes:

$y=-5\pm4\frac{(x+2)}{3}$

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

$=\frac{2*16}{3}\\ =\frac{32}{3}\\ =10.66$