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)

$2 \sqrt{2}$

Answer 2

The correct answer is (A) : $2 \sqrt{2}$
AB:(λ+1)x+λy=4
AC:λx+(1−λ)y+λ=0
Vertex A is on y-axis
⇒x=0

So y=λ4​,y=λ−1λ​
⇒λ4​=λ−1λ​
⇒λ=2
AB:3x+2y=4
AC:2x−y+2=0
⇒A(0,2) Let C(α,2α+2)
Now (Slope of Altitude through C) (−23​)=−1
(α−12α​)(−23​)=−1⇒α=−21​
So C(−21​,1)

Let Equation of tangent be y=mx+2m3​
m2+2m−3=0
⇒m=1,−3
So tangent which touches in first quadrant at T is
T≡(m2a​,m2a​)
≡(23​,3)
⇒CT=(4+4)​=22​