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(C) 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

Vedclass · Accepted Answer

$10$

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(C) 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 $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$ with the general form,we have $a=1, b=\lambda, c=2, h=-\frac{3}{2}, g=\frac{3}{2}, f=-\frac{5}{2}$.Substituting these values into the condition:$(1)(\lambda)(2) + 2(-\frac{5}{2})(\frac{3}{2})(-\frac{3}{2}) - (1)(-\frac{5}{2})^2 - (\lambda)(\frac{3}{2})^2 - (2)(-\frac{3}{2})^2 = 0$$2\lambda + \frac{45}{4} - \frac{25}{4} - \frac{9\lambda}{4} - \frac{18}{4} = 0$$2\lambda - \frac{9\lambda}{4} + \frac{45-25-18}{4} = 0$$-\frac{\lambda}{4} + \frac{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|$.Substituting $a=1, b=2, h=-\frac{3}{2}$:$	an 	heta = \left| \frac{2\sqrt{(-\frac{3}{2})^2 - (1)(2)}}{1+2} ight| = \left| \frac{2\sqrt{\frac{9}{4} - 2}}{3} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} ight| = \frac{2(\frac{1}{2})}{3} = \frac{1}{3}$.Therefore,$\csc^2 	heta = 1 + \cot^2 	heta = 1 + (\frac{1}{	an 	heta})^2 = 1 + (3)^2 = 1 + 9 = 10$.

If $\theta$ is the angle between the lines represented by $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0,$ where $\lambda$ is a real number,then $\csc^2 \theta$ equals:

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