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

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(B) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Here,the parabola is $y^2 = 4x$,so $a = 1$.The equa

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$3$

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(B) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Here,the parabola is $y^2 = 4x$,so $a = 1$.The equation of the tangent is $y = mx + \frac{1}{m}$.Since the tangent passes through the point $(2, 3)$,we substitute these coordinates into the equation:$3 = m(2) + \frac{1}{m}$$3 = 2m + \frac{1}{m}$Multiplying by $m$ (where $m 
eq 0$):$3m = 2m^2 + 1$$2m^2 - 3m + 1 = 0$.This is a quadratic equation in $m$,where $m_1$ and $m_2$ are the roots.From the properties of quadratic equations $am^2 + bm + c = 0$,the sum of the roots is $m_1 + m_2 = -\frac{b}{a}$ and the product is $m_1 m_2 = \frac{c}{a}$.Here,$m_1 + m_2 = \frac{3}{2}$ and $m_1 m_2 = \frac{1}{2}$.We need to find $\frac{1}{m_1} + \frac{1}{m_2} = \frac{m_1 + m_2}{m_1 m_2}$.Substituting the values: $\frac{3/2}{1/2} = 3$.

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point $(2, 3)$ to the parabola $y^2 = 4x$,then what is the value of $\frac{1}{m_1} + \frac{1}{m_2}$?

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