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 :

• 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 $y=\frac{4}{\lambda },y=\frac{\lambda }{\lambda -1}$
$\Rightarrow \frac{4}{\lambda }=\frac{\lambda }{\lambda -1}$
$\Rightarrow \lambda =2$
$AB:3x+2y=4$
$AC:2x-y+2=0$
$\Rightarrow A(0,2)$ 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 $y=mx+\frac{3}{2m}$
$m^2+2m-3=0$
$\Rightarrow m=1,-3$
So tangent which touches in first quadrant at $T$ is
$T\equiv \left(\frac{a}{m^2},\frac{2a}{m}\right)$
$\equiv \left(\frac{3}{2},3\right)$
$\Rightarrow CT=\sqrt{4+4}=2\sqrt{2}$