The equations of the sides AB and AC of a triangle ABC are ( +1) … - Vidyadip

MCQ

The equations of the sides $AB$ and $AC$ of a triangle $ABC$ are $(\lambda+1) x +\lambda y =4 \text { and } \lambda x +(1-\lambda) y +\lambda=0$ 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
$\sqrt{6}$
✓
$2 \sqrt{2}$
C
$2$
D
$4$

✓

Answer

Correct option: B.

$2 \sqrt{2}$

b
$AB :(\lambda+1) x +\lambda y =4$

$AC : \lambda x +(1-\lambda) y +\lambda=0$

Vertex $A$ is on $y$-axis

$\Rightarrow x=0$

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

$\Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1}$

$\Rightarrow \lambda=2$

$AB : 3 x +2 y =4$

$AC : 2 x - y +2=0$

$\Rightarrow A (0,2) \text { Let } C (\alpha, 2 \alpha+2)$

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 $y=m x+\frac{3}{2 m}$

$m ^2+2 m -3=0$

$\Rightarrow m =1,-3$

So tangent which touches in first quadrant at $T$ is

$T \equiv\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)$

$\equiv\left(\frac{3}{2}, 3\right)$

$\Rightarrow CT =\sqrt{4+4}=2 \sqrt{2}$

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