Q.

The equations of the sides AB and AC of a triangle ABC are (λ+1)x+λy=4 and λx+(1-λ)y+λ=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 be part of the parabola y2=6x in the first quadrant is               [2023]

1 22  
2 2  
3 6  
4 4  

Ans.

(1)

AB:(λ+1)x+λy=4

AC:λx+(1-λ)y+λ=0

Since vertex A is on y-axis x=0

So, y=4λ and y=λλ-1
4λ=λλ-1λ=2

AB:3x+2y=4

AC:2x-y+2=0A(0,2)

Let coordinates of C be (α,2α+2).

Now, [slope of altitude through C]×(-32)=-1

(2αα-1)(-32)=-1α=-12

So, coordinates of C are (-12,1).

Equation of tangent be y=mx+32m

  m=1,-3

So, tangent which touches in first quadrant at T is  

T(am2,2am)(32,3)

  CT=(32+12)2+(3-1)2=4+4=22