Question

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 x² + 2xy + 3y² + 4x + 8y – 11 = 0 is tan^–12√2/3.

Answer 1

Equation of the given curve is x² + 2xy + 3y² + 4x + 8y – 11 = 0 

and equation of the given straight line is y – 3x = 2;

Making equation (1) homogeneous equation of the second degree in x any y with the help of (1), we have

or x² + 2xy + 3y² + 1/2  (4xy + 8y² – 12x² – 24 xy) – 11/4 (y² – 6xy + 9x²) = 0 

or 4x² + 8xy + 12y² + 2(8y² – 12x² – 20xy) – 11 (y² – 6xy + 9x²) = 0 

or –119x² + 34xy + 17y² = 0 

or 119x² – 34xy – 17y² = 0 

or 7x² – 2xy – y² = 0 

This is the equation of the lines joining the origin to the points of intersection of (1) and (2). 

Comparing equation (3) with the equation ax² + 2hxy + by 2 = 0 

we have a = 7, b = –1 and 2h = –2 i.e. h = –1 

If θ be the acute angle between pair of lines (3), then