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

\frac{π}{6}

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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 B $\frac{π}{3}$
$4 y = 2 (x + 1)$
$y = (\frac{x + 1}{2})$
${(x + 1)}^{2} = 16 x$
$x^{2} + 2 x + 1 = 16 x$
$x^{2} - 12 x + 1 = 0$
$t y = x + t^{2} (e q u a t i o n o f t a n g e t a t t^{2}, 2 t)$
$t^{2} - 4 t + 1 = 0 \to t_{1} + t_{2} = 4 t_{1} t_{2} = 1$
$m . = \frac{1}{t}$
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$
$= \frac{(t_{2} - t_{1})}{1 + t_{2} t_{1}}$
$\frac{\sqrt{{(t_{1} + t_{2})}_{1}^{2} - 4 t_{1} t_{2}}}{1 + t_{1} t_{2}}$
$\frac{\sqrt{16 - 4}}{1 + 1}$
$\sqrt{3}$
$tan θ = \sqrt{3}$
$θ = π / 3$

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