If the lines represented by (1+sin^2 theta) x | vedclass

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Question

(C) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.If these lines are perpendicular

Vedclass · Accepted Answer

$\frac{3\pi}{2}$

Vedclass · Answer

(C) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.If these lines are perpendicular to each other,the condition is $a + b = 0$.Given the equation $(1+\sin^2 	heta) x^2 + 2hxy + 2\sin 	heta y^2 = 0$,we identify $a = 1+\sin^2 	heta$ and $b = 2\sin 	heta$.Applying the condition $a + b = 0$:$(1+\sin^2 	heta) + 2\sin 	heta = 0$$(1+\sin 	heta)^2 = 0$$1+\sin 	heta = 0$$\sin 	heta = -1$For $	heta \in [0, 2\pi]$,the value of $	heta$ satisfying $\sin 	heta = -1$ is $	heta = \frac{3\pi}{2}$.

If the lines represented by $(1+\sin^2 \theta) x^2+2hxy+2\sin \theta y^2=0$,where $\theta \in [0, 2\pi]$,are perpendicular to each other,then $\theta = \dots$.

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