If {m_1} and {m_2} are the slopes of the tang | vedclass

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(D) The equation of a line passing through $(6, 2)$ with slope $m$ is $y - 2 = m(x - 6)$,which simplifies to $y = mx + (2 - 6m)$.For this line to be a

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Both $(A)$ and $(B)$

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(D) The equation of a line passing through $(6, 2)$ with slope $m$ is $y - 2 = m(x - 6)$,which simplifies to $y = mx + (2 - 6m)$.For this line to be a tangent to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$,the condition of tangency $c^2 = a^2m^2 - b^2$ must be satisfied,where $c = 2 - 6m$,$a^2 = 25$,and $b^2 = 16$.Substituting these values: $(2 - 6m)^2 = 25m^2 - 16$.Expanding the left side: $4 + 36m^2 - 24m = 25m^2 - 16$.Rearranging the terms: $11m^2 - 24m + 20 = 0$.Since ${m_1}$ and ${m_2}$ are the roots of this quadratic equation,by Vieta's formulas:Sum of roots: ${m_1} + {m_2} = -(\frac{-24}{11}) = \frac{24}{11}$.Product of roots: ${m_1}{m_2} = \frac{20}{11}$.Thus,both $(A)$ and $(B)$ are correct.

If ${m_1}$ and ${m_2}$ are the slopes of the tangents to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ which pass through the point $(6, 2)$,then:

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