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+1 vote
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in Mathematics by (65.8k points)

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.

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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

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