Find the centre, foci, eccentricity, vertices of the hyperbola 11x2 - 25y2 - 44x + 50y - 256 = 0 and sketch the graph.
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Answer:
Given equation of hyperbola is x
2
−3y
2
+6x+6y+18=0
(x
2
+6x)−(3y
2
−6y)=−18
(x
2
+6x+9−9)−3(y
2
−2y+1−1)=−18
(x+3)
2
−9−3(y−1)
2
+3=−18
3(y−1)
2
−(x+3)
2
=12
(÷ by 12)
4
(y−1)
2
−
12
(x+3)
2
=1; Centre (−3,1)
Transverse axis is parallel to y-axis
a
2
=4,b
2
=12
e=
1+
a
2
b
2
=
1+
4
12
=
1+3
=2
a=2,e=2
ae=2×2=4;e=2
Centre=(−3,1); Foci=(0,±ae)=(0,±4)
(i.e.,) F
1
=(0,4)+(−3,1)=(−3,5)
F
2
=(0,−4)+(−3,1)=(−3,−5)
Vertices (0,±a)=(0,±2)
(i.e.,) V
1
=(0,2)+(−3,1)=(−3,3)
Step-by-step explanation:
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