Find
(i) The foci of the hyperbola 9x 2 – 16y 2 + 18x + 32y – 151 = 0
(ii) Equation of the hyperbola if vertex and focus of hyperbola are (2, 3) and (6, 3) respectively and eccentricity e of the hyperbola is 2
Text Solution
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(i) (4, 1), (–6, 1)
(ii) 
Sol. (i) The given equation can be written as
9(x 2 + 2x) – 16(y 2 – 2y) = 151 or 9(x 2 + 2x + 1) – 16(y 2 – 2y + 1) = 151 + 9 – 16 = 144
or 
where X = x + 1, Y = y – 1, a 2 = 16, b 2 = 9 ; centre is X = 0, Y = 0 i.e., (–1, 1)
b 2 = a 2 (e 2 – 1) ⇒ e = 5/4
foci are X = ±ae, Y = 0 or x + 1 = ±4 (5/4) ; y – 1 = 0 or (4, 1) and (–6, 1)
directrices X = ±a/e or x + 1 = ±16/5 or 5x – 11 = 0, 5x + 21 = 0.
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