If the lines represented by the equation 6x^2 | vedclass

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Question

(A) The given equation of the pair of lines is $6x^2 + 41xy - 7y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$

Vedclass · Accepted Answer

$-6/7$

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(A) The given equation of the pair of lines is $6x^2 + 41xy - 7y^2 = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$2h = 41$,and $b = -7$.Let the slopes of the two lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the product of the slopes is given by $m_1 m_2 = \frac{a}{b}$.Substituting the values,we get $	an \alpha \cdot 	an \beta = \frac{6}{-7} = -\frac{6}{7}$.

If the lines represented by the equation $6x^2 + 41xy - 7y^2 = 0$ make angles $\alpha$ and $\beta$ with the $x$-axis,then $\tan \alpha \cdot \tan \beta = $

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