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sketch and discuss the hyperbola::x^2-4y+6x+16y-11=0

Posted in Calculus, asked by joseph, 6 years ago. 2464 hits.

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given equation,x^2-4y^2+6x+16y-11=0

 we can rearrange the given equation,x^2+6x-4y^2+16y=11

                                                      x^2+6x-4(y^2-4y)=11

                                                       x^2+6x+9-4(y^2-4y+4)=11+9-16

                                                        (x+3)^2-4(y-2)^2=4

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

equation of hyperbola transverse axis(horizontal)=(x-h)^2/a^2-(y-k)^2/b^2=1

the above equation (1) looking like this equation so,

center of the equation(1) is,(h,k)=(-3,2)

vertices is,(a,0),(-a,0)=(2,0)(-2,0)

foci=(c,0),(-c,0)

c^2= a^2+b^2= 2^2+1^2=5=>c=root5        

foci=(root5,0),(-root5,0)

asymptotes=>y=(+or-)b/ax=>(+or-)1/2x

Venkat Nagendra Thati
Venkat Nagendra Thati - 6 years ago
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