Identify the type of curve which is represented by the equation 2x2+4=4x+y2
Find the foci and vertices(if any), and sketch the graph
y2−2(x2−2x)=4Factor and group termsy2−2(x2−2x+1)=4−2Complete the square; Add (−22)2=1 on the left and subtract 2 from the righty2−2(x−1)2=2Perfect squarey22−(x−1)2=1Divide by 2
The equation is hyperbola that has the form (y−k)2a2−(x−h)2b2=1 with center at (h,k) and vertical transverse axis.
The graph of the shifted hyperbola is obtained from the graph of y22−x2=1 by shifting it 1 unit to the right. This gives us a
a2=2 and b2=1, so a=√2,b=1 and c=√a2+b2=√2+1=√3. Then, by applying transformation
center (h,k)→(1,0)vertices (0,a)→(0,√2)→(0+1,√2)=(1,√2)(0,−a)→(0,−√2)→(0+1,−√2)=(1,−√2)foci (0,c)→(0,√3)→(0+1,√3)=(1,√3)(0,−c)→(0,−√3)→(0+1,−√3)=(1,−√3)
Then, the graph is
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