prove that angle between the lines joining the origin to the points of intersection of straight lines y=3x+2 with the curve x^2+2xy+3y^2+4x+8y-11=0 is tan^-1(0.942)

Question

Vikas TU · Answer

Origin lies on y=3x+2. Taking out x from condition utilizing x = y-2/3 from first condition and utilizing it in x^2+2xy+3y^2+4x+8y-11=0 condition you will get is 34y2+74y-167=0 arrangements are y = 47 and y=121 we realize that x=y-2/3 on substituting , we get so purposes of crossing point are (15, 47) and (119/3 , 121) as should be obvious they lie in various quadrants and source is between them at stake y=3x+2 so point is tan-1(0.942)

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prove that angle between the lines joining the origin to the points of intersection of straight lines y=3x+2 with the curve x^2+2xy+3y^2+4x+8y-11=0 is tan^-1(0.942)

prove that angle between the lines joining the origin to the points of intersection of straight lines y=3x+2 with the curve x^2+2xy+3y^2+4x+8y-11=0 is tan^-1(0.942)

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prove that angle between the lines joining the origin to the points of intersection of straight lines y=3x+2 with the curve x^2+2xy+3y^2+4x+8y-11=0 is tan^-1(0.942)

prove that angle between the lines joining the origin to the points of intersection of straight lines y=3x+2 with the curve x^2+2xy+3y^2+4x+8y-11=0 is tan^-1(0.942)

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