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 $y^2=6 x$ in the first quadrant is :

• A. $2 \sqrt{2}$
• B. 2
• C. $\sqrt{6}$
• D. 4

Answer 1

Answer: (A)

$ AB :(\lambda+1) x +\lambda y =4$
$ AC : \lambda x +(1-\lambda) y +\lambda=0$
Vertex A is on $y$-axis
$\Rightarrow x=0$

So $y =\frac{4}{\lambda}, y =\frac{\lambda}{\lambda-1}$
$ \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} $
$ \Rightarrow \lambda=2$
$A B: 3 x+2 y=4$
$A C: 2 x-y+2=0$
$\Rightarrow A (0,2)$ Let $C (\alpha, 2 \alpha+2)$
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 $C \left(-\frac{1}{2}, 1\right)$

Let Equation of tangent be $y = mx +\frac{3}{2 m }$
$ m^2+2 m-3=0 $
$ \Rightarrow m=1,-3$
So tangent which touches in first quadrant at $T$ is
$T \equiv\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right) $
$ \equiv\left(\frac{3}{2}, 3\right)$
$ \Rightarrow CT =\sqrt{4+4}=2 \sqrt{2}$