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I need this problem from my prep guide to be answered.Subject is pre calculus

I need this problem from my prep guide to be answered.Subject is pre calculus-example-1
User Mohghaderi
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1 Answer

7 votes

since the transverse axis is parallel to the x - axis . we will use :

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

and a = semi congugate axis /2 = 8/2 = 4

and b = semi transverve axis /2 = 12/2 = 6

lets substitute :

( x-h/a)^2 - (y-k/b)^2 = 1

( x- (-3) / 4) - ( y - 2)/6)^2 = 1

((x+3)/4))^2 - (( y-2) /6)^2 = 1

(x+3)^2 / 16 - (y-2)^2 / 36 = 1

This means choose option number 2 for the first box answer, and

choose option number 4 for the second box .

User YourMJK
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