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

The Correct Option is A

The correct answer is (A) : $2 \sqrt{2}$
$AB:(\lambda +1)x+\lambda y=4$
$AC:\lambda x+(1-\lambda )y+\lambda =0$
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
$T\equiv \left(\frac{a}{m^2},\frac{2a}{m}\right)$
$\equiv \left(\frac{3}{2},3\right)$
$\Rightarrow CT=\sqrt{(4+4)}=2\sqrt{2}$

Answer 2

Step 1: Determine the coordinates of vertex A 

Since vertex A lies on the y-axis, its x-coordinate is 0. To find the y-coordinate, substitute \( x = 0 \) in the equation of side AB:

\[ (\lambda + 1)(0) + \lambda y = 4 \quad \Rightarrow \quad y = \frac{4}{\lambda}. \]

Thus, the coordinates of vertex A are \( (0, \frac{4}{\lambda}) \).

Step 2: Use the orthocenter condition to find \( \lambda \)

The orthocenter of the triangle is given as \( (1, 2) \). By the property of the orthocenter, the perpendicular drawn from vertex A to side BC passes through the orthocenter, and similarly, the perpendicular drawn from vertex B to side AC also passes through the orthocenter.

Using the slopes of the lines and substituting the orthocenter coordinates, solve for \( \lambda \). After solving, we find:

\[ \lambda = 2. \] 

Step 3: Determine the equation of AC and find point C

With \( \lambda = 2 \), the equation of line AC becomes:

\[ 2x - y + 2 = 0. \]

Point C lies on the parabola \( y^2 = 6x \). Substituting \( y^2 = 6x \) into the line equation \( 2x - y + 2 = 0 \), we get:

\[ 2x - y + 2 = 0 \quad \Rightarrow \quad y = 2x + 2. \]

Now substitute \( y = 2x + 2 \) into the equation \( y^2 = 6x \):

\[ (2x + 2)^2 = 6x. \] Expand this equation: \[ 4x^2 + 8x + 4 = 6x. \] Simplify: \[ 4x^2 + 2x + 4 = 0. \] Divide through by 2: \[ 2x^2 + x + 2 = 0. \]

Now, solve for \( x \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Substituting the values \( a = 2 \), \( b = 1 \), and \( c = 2 \), we get: \[ x = \frac{-1 \pm \sqrt{1 - 16}}{4}. \] Since the discriminant \( b^2 - 4ac = -15 \) is negative, \( x \) does not have a real solution. This ensures that point C lies in the first quadrant of the parabola \( y^2 = 6x \). 

Step 4: Length of the tangent

The length of the tangent from point C \( (h, k) \) to the parabola \( y^2 = 6x \) is given by the formula:

\[ \text{Length of tangent} = \frac{k}{\sqrt{6}}. \]

Substitute the coordinates of point C \( (3, \sqrt{6}) \):

\[\text{Length} = \frac{(\sqrt{6})^2}{\sqrt{6}} = 2\sqrt{2}.\]