The equations of the sides AB and AC of a triangle ABC are (+1)x+ y=4 and x+(1-)y+=0 respe…

Question

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:

Accepted Answer

$$
\begin{aligned}
& \mathrm{AB}:(\lambda+1) x+\lambda y=4 \\
& \mathrm{AC}: \lambda x+(1-\lambda) y+\lambda=0 \\
& \text { Vertex } A \text { is on } y \text {-axis } \\
& \Rightarrow x=0
\end{aligned}
$$

So $\mathrm{y}=\frac{4}{\lambda}, \mathrm{y}=\frac{\lambda}{\lambda-1}$

$$
\begin{aligned}
& \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} \\
& \Rightarrow \lambda=2 \\
& \mathrm{AB}: 3 x+2 y=4 \\
& \mathrm{AC}: 2 \mathrm{x}-\mathrm{y}+2=0 \\
& \Rightarrow \mathrm{A}(0,2) \text { Let } \mathrm{C}(\alpha, 2 \alpha+2)
\end{aligned}
$$

Now (Slope of Altitude through C) $\left(-\frac{3}{2}\right)=-1$

$$
\left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2}
$$

So $\mathrm{C}\left(-\frac{1}{2}, 1\right)$

Let Equation of tangent be $y=m x+\frac{3}{2 m}$

$$
\begin{aligned}
& \mathrm{m}^2+2 \mathrm{~m}-3=0 \\
& \Rightarrow \mathrm{m}=1,-3
\end{aligned}
$$

So tangent which touches in first quadrant at $\mathrm{T}$ is

$$
\begin{aligned}
& \mathrm{T} \equiv\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right) \\
& \equiv\left(\frac{3}{2}, 3\right) \\
& \Rightarrow \mathrm{CT}=\sqrt{4+4}=2 \sqrt{2}
\end{aligned}
$$

Solutions · 1