The angle between the lines represented by th | vedclass

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Question

(C) The given equation is $4x^2 - 24xy + 11y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = -24 \Rightarro

Vedclass · Accepted Answer

$	an^{-1}\frac{4}{3}, 	an^{-1}\left(-\frac{4}{3}ight)$

Vedclass · Answer

(C) The given equation is $4x^2 - 24xy + 11y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = -24 \Rightarrow h = -12$,and $b = 11$.The angle $	heta$ between the pair of lines is given by the formula $	an 	heta = \pm \frac{2\sqrt{h^2 - ab}}{a + b}$.Substituting the values: $	an 	heta = \pm \frac{2\sqrt{(-12)^2 - (4)(11)}}{4 + 11} = \pm \frac{2\sqrt{144 - 44}}{15} = \pm \frac{2\sqrt{100}}{15} = \pm \frac{2 	imes 10}{15} = \pm \frac{20}{15} = \pm \frac{4}{3}$.Therefore,$	heta = 	an^{-1}\left(\pm \frac{4}{3}ight)$,which means $	heta = 	an^{-1}\frac{4}{3}$ or $	heta = 	an^{-1}\left(-\frac{4}{3}ight)$.

The angle between the lines represented by the equation $4x^2 - 24xy + 11y^2 = 0$ is given by:

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