What is the angle between the tangents drawn | vedclass

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(B) The equation of any tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.Since the tangent passes through $(1, 4)$,we have $4 = m(1) + \fr

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$\frac{\pi}{3}$

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(B) The equation of any tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.Since the tangent passes through $(1, 4)$,we have $4 = m(1) + \frac{1}{m}$.Multiplying by $m$,we get $m^2 - 4m + 1 = 0$.Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1 + m_2 = 4$ and $m_1m_2 = 1$.The difference between the slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2} = \sqrt{4^2 - 4(1)} = \sqrt{12} = 2\sqrt{3}$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{2\sqrt{3}}{1 + 1} ight| = \sqrt{3}$.Therefore,$	heta = \frac{\pi}{3}$.

What is the angle between the tangents drawn from $(1, 4)$ to the parabola $y^2 = 4x$?

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