Two tangents are drawn from point $(1, 4)$ to the parabola $y^{2} = 4 x .$ Angles between tangents is -

\frac{π}{ϵ}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{3}$
Any tangent to $y^{2} = 4 x$ is $y = m x + \frac{1}{m}$ .

It passes through $(1, 4)$

$\Rightarrow 4 = m + \frac{1}{m}$

$\Rightarrow m^{2} - 4 m + 1 = 0$

Let $m_{1}, m_{2}$ be the roots of the quadratic equation then

$\Rightarrow m_{1} + m_{2} = 4$ and $m_{1} m_{2} = 1$

$\Rightarrow {(m_{1} - m_{2})}^{2} = {(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}$

$\Rightarrow {(m_{1} - m_{2})}^{2} = {(4)}^{2} - 4 \times 1$

$\Rightarrow {(m_{1} - m_{2})}^{2} = 16 - 4 = 12$

$\Rightarrow | m_{1} - m_{2} | = \sqrt{12} = 2 \sqrt{3}$

Now, $tan θ = \frac{| m_{1} - m_{2} |}{1 + m_{1} m_{2}} = \frac{2 \sqrt{3}}{1 + 1} = \frac{2 \sqrt{3}}{2} = \sqrt{3}$

$\Rightarrow tan θ = \sqrt{3}$

$\Rightarrow θ = \frac{π}{3}$

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