Two tangents are drawn from point $(1, 4)$ to the parabola $y^{2} = 4 x .$ Angles between tangents 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 A $\frac{π}{3}$
Equation of a line through point $(1, 4)$ IS
$(y - 4) = m (x - 1)$
let $m$ is slope of line
it is a tangent to parabola
$y^{2} = 4 x$
$\Rightarrow$ Solving
$y^{2} = 4 (1 + \frac{y}{m} - \frac{4}{m})$
$\Rightarrow y^{2} = 4 + \frac{4 y}{m} - \frac{16}{m}$
$\Rightarrow y^{2} - \frac{4 y}{m} + (\frac{16}{m} - 4) = 0$
The equation will have equal rests (As tangent touch parabola at only one point)
$\Rightarrow D = 0$
$\frac{16}{m^{2}} - 4 (\frac{16}{m} - 4) = 0$
$\frac{1}{m^{2}} - (\frac{4}{m} - 1) = 0$
$\frac{1}{m^{2}} - \frac{4}{m^{2}} + 1 = 0 \Rightarrow m^{2} - 4 m + 1 = 0$
The above equation will have two values of $m$ ( $m_{1}$ & $m_{2}$ ) $\Rightarrow$ two tangents $m_{1} + m_{2} = 4, m_{1} m_{2} = 1$
$\Rightarrow (m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} + m_{2} = 16 - 4 = 12$
$\Rightarrow m_{1} - m_{2} = 2 \sqrt{3}$
Now $tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{m_{1} + m_{2}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{2 \sqrt{3}}{1 + 1} ∣ ∣ ∣$
$= \frac{2 \sqrt{3}}{2} \Rightarrow tan θ = \sqrt{3}$
$\Rightarrow θ = \frac{π}{3}$

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