If the line y = 2x + lambda is a tangent to t | vedclass

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(C) The equation of the hyperbola is $36x^2 - 25y^2 = 3600$. Dividing by $3600$,we get $\frac{x^2}{100} - \frac{y^2}{144} = 1$.Here,$a^2 = 100$ and $b

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$\pm 16$

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(C) The equation of the hyperbola is $36x^2 - 25y^2 = 3600$. Dividing by $3600$,we get $\frac{x^2}{100} - \frac{y^2}{144} = 1$.Here,$a^2 = 100$ and $b^2 = 144$.The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given $m = 2$ and $c = \lambda$,we have $\lambda^2 = (100)(2^2) - 144$.$\lambda^2 = 400 - 144 = 256$.Therefore,$\lambda = \pm \sqrt{256} = \pm 16$.

If the line $y = 2x + \lambda$ is a tangent to the hyperbola $36x^2 - 25y^2 = 3600$,then $\lambda = $

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