The equations of the sides $AB$ and $AC$ of 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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$\sqrt{6}$

$2 \sqrt{2}$

answer is B.

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Detailed Solution

$AB : (λ + 1) x + λ y = 4$

$AC : λ x + (1 - λ) y + λ = 0$

Vertex $A$ is on $y$ -axis
$\Rightarrow x = 0$
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So $y = \frac{4}{λ}, y = \frac{λ}{λ - 1}$
$\Rightarrow \frac{4}{λ} = \frac{λ}{λ - 1}$ $\Rightarrow λ = 2$
$A B : 3 x + 2 y = 4$
$AC : 2 x - y + 2 = 0$
$\Rightarrow A (0, 2)$ Let $C (α, 2 α + 2)$
Now (Slope of Altitude through C) $(- \frac{3}{2}) = - 1$
$(\frac{2 α}{α - 1}) (- \frac{3}{2}) = - 1 \Rightarrow α = - \frac{1}{2}$
So $C (- \frac{1}{2}, 1)$
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