SOLUTION: can you please graph the hyerbola and explain how you did it (x+2)^2/4 - (y-4)^2/25=1 thank you!!

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Question 199588: can you please graph the hyerbola and explain how you did it
(x+2)^2/4 - (y-4)^2/25=1
thank you!!

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
can you please graph the hyerbola and explain how you did it
%28x%2B2%29%5E2%2F4+-+%28y-4%29%5E2%2F25=1
thank you!!

%28%28x-h%29%5E2%29%2Fa%5E2+-+%28%28y-k%29%5E2%29%2Fb%5E2+=+1
 
h+=+-2, k=4, 
 
a%5E2=4, so a=2
 
b%5E2=25, so b=5
 
The center (h,k) = (-2,4)
 
We start out plotting the center C(h,k) = C(-2,4)
 

 
Next we draw the left semi-transverse axis,
which is a segment a=2 units long horizontally 
left from the center.  This semi-transverse
axis ends up at one of the two vertices (-4,4).
We'll call it V1(-4,4):
 

 
Next we draw the right semi-transverse axis,
which is a segment a=5 units long horizontally 
right from the center. This other semi-transverse
axis ends up at the other vertex (0,4).
We'll call it V2(0,4):
 

 
That's the whole transverse ("trans"="across",
"verse"="vertices", the line going across from
one vertex to the other. It is 2a in length,
so the length of the transverse axis is 2a=2(2)=4
 
Next we draw the upper semi-conjugate axis,
which is a segment b=5 units long verically 
upward from the center.  This semi-conjugate
axis ends up at (-2,9).


 

 
Next we draw the lower semi-conjugate axis,
which is a segment b=8 units long verically 
downward from the center.  This semi-conjugate
axis ends up at (-2,-1). 
 


 
That's the complete conjugate axis. It is 2b in length,
so the length of the transverse axis is 2b=2(5)=10
 
Next we draw the defining rectangle which has the
ends of the transverse and conjugate axes as midpoints
of its sides:



 
Next we draw and extend the two diagonals of this defining
rectangle:



Now we can sketch in the hyperbola:


 
Edwin