If the line 5x+12y=9 touches the hyperbola x2−9y2=9, then its point of contact is
A
(5,−43)
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B
(5,43)
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C
(−3,2)
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D
(3,0)
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Solution
The correct option is A(5,−43) Given line line 5x+12y=9 or y=−5x12+34
any tangent to the given hyperbola is y=mx±√9m2−1
on comparing with given equation m=−512,c=34
Hence point of contact will be (−a2mc,−b2c)≡(5,−43)