To find: The equation of the given hyperbola.The foci of the hyperbola is (0, pm 5) and eccentricity is e-1.5.

UkusakazaL 2020-11-23 Answered

To find: The equation of the given hyperbola. The foci of the hyperbola is (0, ± 5)and entricity is e1.5.

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l1koV
Answered 2020-11-24 Author has 100 answers
Step 1 Write the general equation of the hyperbola. (y  k)2a2  (x  h)2b2=1 The center of the hyperbola is (h, k), the vertice are (h ± a,k), and foci are (h ± c,k). Write the expression for c. c=a2 + b2 Write the expression for the eccentricity of the hyperbola. e=a2 + b2a=ca The foci of the conic section are, (h,k ± c)=(0, ± 5)h=0c=5k=0 Step 2 Write the expression for the eccentricity of the hyperbola. e=ca Substitute 5 for c and 1.5 for e. 1.5=5a
a=3.33 Write the expression for c. c=a2 + b2 Substitute 3.33 for a and 5 for c. 5=3.332 + b2
25=3.332 + b2
13.91=b2
b=3.73 Write the general equation of the hyperbola. (y  k)2a2  (x  h)2b2=1 Substitute 3.33 for a, 0 for h, 3.73 for b, and 0 for k. (y  0)23.332  (x  0)23.732=1
y211.08  x213.91=1 Therefore, the equation of the given hyperbola is y211.08  x213.91=1.
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