The angle between the tangents drawn from the | vedclass

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

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$\frac{\pi}{3}$

Vedclass · Answer

(C) The equation of any tangent to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$.Since the tangents pass through the point $(1,4)$,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 $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.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| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.Therefore,$	heta = 	an^{-1}(\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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