If m_1 and m_2 are the slopes of the tangents | vedclass

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx + \frac{a}{m}$.Here,$4a = 11$,so $a = \frac{11}{4}$.The t

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$21$

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx + \frac{a}{m}$.Here,$4a = 11$,so $a = \frac{11}{4}$.The tangent passes through the point $(1, 4)$,so $4 = m(1) + \frac{11}{4m}$.Multiplying by $4m$,we get $16m = 4m^2 + 11$,which simplifies to $4m^2 - 16m + 11 = 0$.Since $m_1$ and $m_2$ are the roots of this quadratic equation,we have $m_1 + m_2 = \frac{16}{4} = 4$ and $m_1m_2 = \frac{11}{4}$.We need to find $2(m_1^2 + m_2^2) = 2((m_1 + m_2)^2 - 2m_1m_2)$.Substituting the values,we get $2(4^2 - 2 	imes \frac{11}{4}) = 2(16 - \frac{11}{2}) = 2(\frac{32 - 11}{2}) = 21$.

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 11x$,then $2(m_1^2 + m_2^2) = $

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