The equations of the sides AB and AC of a triangle ABC are $(\lambda+1)x + \lambda y = 4$ and $\lambda x + (1-\lambda)y + \lambda = 0$ 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 $y^2 = 6x$ in the first quadrant is
Step-by-Step Solution
Key Concept: Find vertex A (on y-axis), use orthocentre condition to find $\lambda$, locate C, then find tangent from C to parabola
Setting $x=0$ in AB gives $A(0,2)$. Using orthocentre $(1,2)$ and perpendicularity conditions, $\lambda=2$. Lines: AB: $3x+2y=4$, AC: $2x-y+2=0$. C lies on AC with slope condition from orthocentre giving $\alpha = -1/2$, so $C(-1/2, 1)$. Tangent from $C(-1/2,1)$ to $y^2=6x$ (i.e., $a=3/2$): tangent $y = mx + 3/(2m)$. Passes through $C$: $1 = -m/2 + 3/(2m) \Rightarrow 2m = -m^2 + 3 \Rightarrow m^2+2m-3=0 \Rightarrow m=1$ (first quadrant). Point $T = (3/2, 3)$. $CT = \sqrt{(3/2+1/2)^2+(3-1)^2} = \sqrt{4+4} = 2\sqrt{2}$. Answer: (2)
Correct Answer: $2\sqrt{2}$