For all real values of m,the straight line y | vedclass

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(D) The equation of the line is $y = mx + \sqrt{9m^2 - 4}$.Squaring both sides,we get $(y - mx)^2 = 9m^2 - 4$.$y^2 - 2mxy + m^2x^2 = 9m^2 - 4$.Rearran

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$4x^2 - 9y^2 = 36$

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(D) The equation of the line is $y = mx + \sqrt{9m^2 - 4}$.Squaring both sides,we get $(y - mx)^2 = 9m^2 - 4$.$y^2 - 2mxy + m^2x^2 = 9m^2 - 4$.Rearranging as a quadratic in $m$: $m^2(x^2 - 9) - 2mxy + (y^2 + 4) = 0$.Since the line is a tangent,the discriminant $D$ must be zero for the quadratic in $m$.$D = (-2xy)^2 - 4(x^2 - 9)(y^2 + 4) = 0$.$4x^2y^2 - 4(x^2y^2 + 4x^2 - 9y^2 - 36) = 0$.$4x^2y^2 - 4x^2y^2 - 16x^2 + 36y^2 + 144 = 0$.$-16x^2 + 36y^2 = -144$.Dividing by $-4$,we get $4x^2 - 9y^2 = 36$.

For all real values of $m$,the straight line $y = mx + \sqrt{9m^2 - 4}$ is a tangent to the curve:

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