How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $(x+1)^2/4-(y+3)^2/9=1$ ?

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Answer 1 · 2018-03-25T13:03:00

The vertices are $(-3,-3) and (1,-3)$ , focii are
$((-1-sqrt13),-3) and ((-1+sqrt13),-3)$ asymptotes
are $y=3/2x-3/2 and y= -3/2x-9/2$

$(x+1)^2/4- (y+3)^2/9=1; h=-1,k=-3,a=2 , b= 3$

This is standard form of the equation of a hyperbola with center

$(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :.$ Centre $(-1,-3)$

The vertices are $a$ units from the center, and the foci are

$c$ units from the center. Moreover $c^2=a^2+b^2$

c $((-1-2),-3) and ((-1+2,3)$ or

$(-3,-3) and (1,-3) ; c^2= 2^2+3^2=13$

$:. c= +- sqrt13 :.$ Focii are at

$((-1-sqrt13),-3) and ((-1+sqrt13),-3)$

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions $2a=4 and 2b=6$ with its

centre at $(-1,-3) :.$ slope $+-b/a=+- 3/2$ Equation of

asymptotes are $y+3= +-3/2(x+1) or y=-3+-3/2(x+1)$

or $y=3/2x-3/2 and y= -3/2x-9/2$ [Ans]

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (x+1)^2/4-(y+3)^2/9=1(x+1)^2/4-(y+3)^2/9=1?