The angle between the lines represented by th | vedclass

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Question

(A) The given equation is $x^2 + 2xy \sec 	heta + y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$h = \sec

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$	heta$

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(A) The given equation is $x^2 + 2xy \sec 	heta + y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$h = \sec 	heta$,and $b = 1$.Let $\alpha$ be the angle between the lines. The formula for the angle is $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an \alpha = \left| \frac{2\sqrt{\sec^2 	heta - 1}}{1 + 1} ight|$.Since $\sec^2 	heta - 1 = 	an^2 	heta$,we have $	an \alpha = \left| \frac{2\sqrt{	an^2 	heta}}{2} ight| = |	an 	heta|$.Therefore,$\alpha = 	heta$.

The angle between the lines represented by the equation $x^2 + 2xy \sec \theta + y^2 = 0$ is

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