The equation x^2-3xy+lambda y^2+3x-5y+2=0,whe | vedclass

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(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - c

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$\sqrt{10}$

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(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Comparing with $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,we have $a=1, h=-3/2, b=\lambda, g=3/2, f=-5/2, c=2$. Substituting these values: $(1)(\lambda)(2) + 2(-5/2)(3/2)(-3/2) - (1)(-5/2)^2 - (\lambda)(3/2)^2 - (2)(-3/2)^2 = 0$. $2\lambda + 45/4 - 25/4 - 9\lambda/4 - 9/2 = 0$. $2\lambda - 9\lambda/4 + 5 - 4.5 = 0 \implies -\lambda/4 + 0.5 = 0 \implies \lambda = 2$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Here $a=1, b=2, h=-3/2$. $	an 	heta = \left| \frac{2\sqrt{9/4 - 2}}{1+2} ight| = \frac{2\sqrt{1/4}}{3} = \frac{1}{3}$. Since $	an 	heta = 1/3$,then $\cot 	heta = 3$. $\operatorname{cosec}^2 	heta = 1 + \cot^2 	heta = 1 + 3^2 = 10$. Thus,$\frac{\operatorname{cosec}^2 	heta}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}$.

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$,where $\lambda$ is a real number,represents a pair of lines. If $\theta$ is the acute angle between the lines,then $\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}} = $

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