The angle between the tangents drawn from the | vedclass

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

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$\frac{\pi}{3}$

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(B) The equation of any tangent to the parabola ${y^2} = 4x$ is given by $y = mx + \frac{1}{m}$.Since the tangent passes through the point $(1, 4)$,we substitute these coordinates into the equation:$4 = m(1) + \frac{1}{m}$Multiplying by $m$,we get the quadratic equation:${m^2} - 4m + 1 = 0$Let the roots be ${m_1}$ and ${m_2}$. From the properties of quadratic equations:${m_1} + {m_2} = 4$ and ${m_1}{m_2} = 1$.The difference between the slopes is given by:$|{m_1} - {m_2}| = \sqrt{({m_1} + {m_2})^2 - 4{m_1}{m_2}} = \sqrt{4^2 - 4(1)} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.If $	heta$ is the angle between the tangents,then:$	an 	heta = \left| \frac{{m_1} - {m_2}}{1 + {m_1}{m_2}} ight| = \frac{2\sqrt{3}}{1 + 1} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Therefore,$	heta = \arctan(\sqrt{3}) = \frac{\pi}{3}$.

The angle between the tangents drawn from the point $(1, 4)$ to the parabola ${y^2} = 4x$ is

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