Prove that the angle between the lines joining the origin to the points of intersection of the straight line `y=3x+2` with the curve `x^2+2x y+3y^2+4x+8y-11=0` is `tan^(-1)((2sqrt(2))/3)`

Options

A. `(sqrt2)/(3)`
B. `(2sqrt2)/(3)`
C. `(2)/(3)`
D. `2sqrt2`

Answer

B

Explanation

The equation of the given straight line is
`y=3x+2rArr(y-3x)/(2)=1" …(i)"`
The equation of the given curve is
`x^(2)+2xy+3y^(2)+4x+8y-11=0" …(ii)"`
The combined equation of the straight lines joining the origin to the points of intersection of (i) and (ii) is a homogeneous equation of second degree obtained with the help of (i) and (ii). Making the equation (ii) homogeneous of the second degree in x and y with the help of (i), we get
`x^(2)+2xy+3y^(2)+4x((y-3x)/(2))+8y((y-3x)/(2))-11((y-3x)/(2))^(2)=0`
`rArr7x^(2)-2xy-y^(2)=0`
This is the required equation. Comparing this equation with `ax^(2)+2hxy+by^(2)=0`, we obtain
`a=7, b=-1 and h=-1`
Let `theta` be the required angle. Then,
`tan theta=(2sqrt(h^(2)-ab))/(a+b)`
`rArr" "tantheta=(2sqrt(1+7))/(7-1)=(2sqrt2)/(3)rArrtheta=tan^(-1)((2sqrt2)/(3))`