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Mathematics-2017-->View question

Asked On2018-01-30 14:17:24 by:Purnima

Taged users:
Aparna-Dasgupta

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Equation of hyperbola is
(x^2/a^2)-(y^2/b^2) = 1
foci is (±2, 0) hence ae = 2, thus, a^2e^2 = 4
b2 = a2(e2 – 1)
a^2 + b^2 = 4 ...(1)
Hyperbola passes through 2^(1/2), 3^(1/2)
2/(a^2)-3/(b^2)=1...(2)
On solving (1) and (2)
a^2 = 8 (is rejected) and a^2 = 1 and b^2 = 3
(x^2)/1-(y^2)/3=1

Equation of tangent is (2^(1/2)x)/1-(3^(1/2)y)/3=1
Hence (2(2)^(1/2),3(3)^(1/2)) satisfy it.

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