Parabola Question 1
Question 1 - 24 January - Shift 2
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
(1) $\sqrt{6}$
(2) $2 \sqrt{2}$
(3) 2
(4) 4
Show Answer
Answer: (2)
Solution:
Formula: Orthocentre, Condition for Perpendicular Lines, Equation of the tangent at any point
$AB:(\lambda+1) x+\lambda y=4$
$AC: \lambda x+(1-\lambda) y+\lambda=0$
Vertex $A$ is on $y$-axis
$\Rightarrow x=0$
So $y=\frac{4}{\lambda}, y=\frac{\lambda}{\lambda-1}$
$\Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1}$
$\Rightarrow \lambda=2$
$AB: 3 x+2 y=4$
$AC: 2 x-y+2=0$
$\Rightarrow A(0,2)$
Let $C(\alpha, 2 \alpha+2)$
Now (Slope of Altitude through C) $(-\frac{3}{2})=-1$
$(\frac{2 \alpha}{\alpha-1})(-\frac{3}{2})=-1 \Rightarrow \alpha=-\frac{1}{2}$
So $C(-\frac{1}{2}, 1)$
