The angle between the tangents drawn to the p | vedclass

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(B) The equation of a tangent to the parabola $y^2=4ax$ (where $a=1$) is $y=mx+\frac{a}{m}$,which simplifies to $y=mx+\frac{1}{m}$.Since the tangent p

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$\frac{\pi}{3}$

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(B) The equation of a tangent to the parabola $y^2=4ax$ (where $a=1$) is $y=mx+\frac{a}{m}$,which simplifies to $y=mx+\frac{1}{m}$.Since the tangent passes through the point $(1,4)$,we substitute these coordinates into the equation:$4 = m(1) + \frac{1}{m}$$4 = m + \frac{1}{m}$$m^2 - 4m + 1 = 0$Let the slopes of the two tangents be $m_1$ and $m_2$. From the quadratic equation,we have:$m_1 + m_2 = 4$ and $m_1 m_2 = 1$.The angle $	heta$ between the two tangents is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$:$(m_1 - m_2)^2 = (4)^2 - 4(1) = 16 - 4 = 12$$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$Therefore,$	an 	heta = \left| \frac{2\sqrt{3}}{1 + 1} ight| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,we have $	heta = \frac{\pi}{3}$.

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

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