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If the line `y=3x+lambda` touches the hyperbola `9x^(2)-5y^(2)=45`, then `lambda`=A. `+-3sqrt(6)`B. `+-6`C. `+-3`D. `+-4` |
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Answer» The equation of the hyperbola is `9x^(2)-5y^(2)=45implies(x^(2))/(5)-(y^(2))/(9)=1`……`(ii)` This of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, where `a^(2)=5` and` b^(2)=9`. The line `y=3x+lambda` will touch the given hyperbola, if `lambda^(2)=5(3)^(2)-9` [Using : `c^(2)=a^(2)m^(2)-b^(2)`] `implieslambda^(2)=36implieslambda=+-6` |
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