Find 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)+2xy+3y^(2)+4x+8y=11=0`.

Explanation

The equation of straight line is
`(y-3x)/(2)=1`
Making given equation of curve homogenous , we get
`x^(2)+2xy+3y^(2)+(4x+8y)((y-3x)/(2))-11((y-3x)/(2))^(2)=0`
or `7x^(2)-2xy-y^(2)=0`
This is the equation of the pair of lines joining origen to the point of intersection of the given line and curve . Comparing this equation with equation with the equation `ax^(2)+2hxy+by^(2)=0`, we get
`a=7,b=-1andh=-1`
If `theta` is the acute angle between component lines of (1) , then
`tantheta=|2sqrt(h^(2)-ab/(a+b))|=|(2sqrt(1+7))/(7-1)|=(2sqrt(8))/(6)=(2sqrt(2))/(3)`
`theta="tan"^(-1)(2sqrt(2))/(3)`