The equation of one of the lines represented | vedclass

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(C) The given equation is $x^2 + 2xy \cot 	heta - y^2 = 0$.This is a homogeneous equation of degree $2$ of the form $ax^2 + 2hxy + by^2 = 0$,where $a

Vedclass · Accepted Answer

$x \sin 	heta + y(\cos 	heta + 1) = 0$

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(C) The given equation is $x^2 + 2xy \cot 	heta - y^2 = 0$.This is a homogeneous equation of degree $2$ of the form $ax^2 + 2hxy + by^2 = 0$,where $a = 1$,$h = \cot 	heta$,and $b = -1$.The lines represented by this equation are given by $y = \frac{h \pm \sqrt{h^2 - ab}}{b} x$.Substituting the values: $y = \frac{\cot 	heta \pm \sqrt{\cot^2 	heta - (1)(-1)}}{-1} x$.$y = -(\cot 	heta \pm \sqrt{\csc^2 	heta}) x$.$y = -(\cot 	heta \pm \csc 	heta) x$.For the positive sign: $y = -(\frac{\cos 	heta + 1}{\sin 	heta}) x \Rightarrow x \sin 	heta + y(\cos 	heta + 1) = 0$.For the negative sign: $y = -(\frac{\cos 	heta - 1}{\sin 	heta}) x \Rightarrow x \sin 	heta + y(\cos 	heta - 1) = 0$.Thus,one of the lines is $x \sin 	heta + y(\cos 	heta + 1) = 0$.

The equation of one of the lines represented by the equation $x^2 + 2xy \cot \theta - y^2 = 0$ is

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