The equations of the sides AB and AC a triangle ABC are $(λ + 1) x + λ y = 4$ and $λ x + (1 - λ) y + λ = 0$ respectively. Its vertex A is on the y-axis and its orthocentre is (1,2).The length of the tangent from the point C to the part of the parabola $y^{2} = 6 x$ in the first
quadrant is

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$2 \sqrt{2}$

$\sqrt{6}$

answer is C.

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Detailed Solution

Let $A = (0, y_{1})$
$λ y_{1} = 4$
And $(1 - λ) y_{1} = - λ$
From (1) & (2)
$\begin{array}{l} y_{1} = 2, A = (0, 2), λ = 2 \\ \bar{AB} = 3 x + 2 y = 4, \bar{AC} = 2 x - y + 2 = 0 \end{array}$
Let $C = [h, 2 h + 2]$
$\begin{array}{l} m_{1} \cdot m_{2} = - 1 \\ \Rightarrow (\frac{- 3}{2}) (\frac{2 h}{h - 1}) = - 1 \\ \Rightarrow h = \frac{- 1}{2} \Rightarrow C = (- \frac{1}{2}, 1) \end{array}$
Equation of tangent is $y = mx + \frac{a}{m} \Rightarrow 1 = m (\frac{- 1}{2}) + \frac{3}{2 m}$
$\begin{array}{c} \Rightarrow m = 1, - 3 \\ C = (\frac{- 1}{2}, 1) \\ P = [\frac{a}{m^{2}}, \frac{2 a}{m}] = (\frac{3}{2}, 3) \\ CP = 2 \sqrt{2} \end{array}$