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(C) The given equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing with $x^

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The given equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ 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 $a=1, b=\lambda, c=2, h=-3/2, g=3/2, f=-5/2$.Substituting these values into the condition:$(1)(\lambda)(2) + 2(-5/2)(3/2)(-3/2) - 1(-5/2)^2 - \lambda(3/2)^2 - 2(-3/2)^2 = 0$$2\lambda + 45/4 - 25/4 - 9\lambda/4 - 18/4 = 0$$2\lambda - 9\lambda/4 + (45-25-18)/4 = 0$$-\lambda/4 + 2/4 = 0 \Rightarrow \lambda = 2$.Now,the angle $	heta$ between the lines is given by $	an	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.$	an	heta = \left| \frac{2\sqrt{(-3/2)^2 - (1)(2)}}{1+2} ight| = \left| \frac{2\sqrt{9/4 - 2}}{3} ight| = \left| \frac{2\sqrt{1/4}}{3} ight| = \frac{2(1/2)}{3} = 1/3$.Thus,$	ext{cosec}^2	heta = 1 + \cot^2	heta = 1 + (1/	an	heta)^2 = 1 + (3)^2 = 1 + 9 = 10$.

If $\theta$ is the angle between the two lines represented by $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,then what is the value of $\text{cosec}^2\theta$? (where $\lambda$ is a real number.)

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