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If the line $y=2 x+\lambda$ be a tangent to the hyperbola $36 x^2-25 y^2=3600$, then $\lambda=$
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The correct answer is:
$\pm 16$
If $y=2 x+\lambda$ is tangent to given hyperabola, then
$\lambda= \pm \sqrt{a^2 m^2-b^2}= \pm \sqrt{(100)(4)-144}= \pm \sqrt{256}= \pm 16$
$\lambda= \pm \sqrt{a^2 m^2-b^2}= \pm \sqrt{(100)(4)-144}= \pm \sqrt{256}= \pm 16$
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