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प्रश्न
Find the eccentricity and the coordinates of foci of the hyperbola 25x2 + 9y2 = 225.
बेरीज
उत्तर
25x2 - 9y2 = 225
The given equation can be written as
`(x^2)/(9) – (y^2)/(25)` = 1
Here, a2 = 9 ⇒ a = 3 and b2 = 25 ⇒ b = 5
Eccentricity,
e = `sqrt(1+(b^2)/(a^2))`
= `sqrt(1+(25)/(9)`
= `sqrt(34)/(9)`
= `sqrt(34)/(3)`
Foci = (±ae,0)
= `(±3 xx sqrt(34)/(3),0)`
= `(±sqrt(34),0)`
Hence, the eccentricity and foci of the given hyperbola are `sqrt(34)/(3) "and" (±sqrt(34), 0)` respectively.
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