Convert the equation to standard form by completing the square on x and y. Then

Question

Convert the equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. Answer parts A-D below.

x^2-y^2-4x-12y-33=0

A. find the standard form

B. Next, find the asymptotes.

C.Next, graph the hyperbola. Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes.

Which of the following is the graph of the hyperbola?

D.Find the foci, located at

left parenthesis h plus c comma k right parenthesis(h+c, k)

and

left parenthesis h minus c comma k right parenthesis(hc, k).

Find c using the equation .c2=a2+b2.

(Simplify your answer. Type an exact answer, using radicals as needed.)

Explanation / Answer

x^2-y^2-4x-12y-33=0

(x^2 -4x) -(y^2 +12y) -33 =0

(x-2)^2 -4 -(y+6)^2 + 36 -33 =0

(x-2)^2 /1 -(y+6)^2/1 =1

standard form of equation with centre ( 2, -6)

foci: (h+c, -k) and (h-c, k)

a = 1 ; b=1 ; c= a^2 +b^2 = sqrt(1+1) = sqrt2

foci (2+sqrt2 , -6) and (2-sqr6 , -6)

Asymtotes: y = +/-b/a(x-h) +k

y = +/- (x- 2) -6

y = x-2-6

y = x-8

; y = -x +2 -6

y = -x -4

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