For all real permissible values of m, if the straight line y=mx+√9m2−4 is tangent to a hyperbola, then equation of the hyperbola can be
A
9x2−4y2=64
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B
4x2−9y2=64
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C
9x2−4y2=36
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D
4x2−9y2=36
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Solution
The correct option is D4x2−9y2=36 Let the equation of tangent to the hyperbola x2a2−y2b2=1 is y=mx±√a2m2−b2 Comparing it with y=mx+√9m2−4, we have a2=9 and b2=4 ∴ Equation of the hyperbola is : x29−y24=1 ⇒4x2−9y2=36