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## A hyperbola passes through the point P(2^(1/2),3^(1/2)) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point :-JEE MAINS-Mathematics-2017

(1) (-2^(1/2),-3^(1/2))

(2) 3x(2)^(1/2),2x3^(1/2)

(3) 2x2^(1/2),3x3^(1/2)

(4) 3^(1/2),2^(1/2)

By:Purnima

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|Aparna-Dasgupta

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## Answers

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.

leo

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