ax^2-4xy-2y^2=0 represents a pair of lines. I | vedclass

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(A) The equation of the pair of lines is $ax^2-4xy-2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=-4 \Rightarrow H=-2$,and $B=-2$.T

Vedclass · Accepted Answer

$11$

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(A) The equation of the pair of lines is $ax^2-4xy-2y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=-4 \Rightarrow H=-2$,and $B=-2$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^2-AB}}{A+B} ight|$.Substituting the values,$	an 	heta = \left| \frac{2\sqrt{(-2)^2 - a(-2)}}{a-2} ight| = \left| \frac{2\sqrt{4+2a}}{a-2} ight|$.Given $\cos 	heta = \frac{1}{5}$,we have $\sin 	heta = \sqrt{1-\cos^2 	heta} = \sqrt{1-\frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{24}/5}{1/5} = \sqrt{24}$.Equating the two expressions for $	an 	heta$: $\frac{2\sqrt{4+2a}}{|a-2|} = \sqrt{24}$.Squaring both sides: $\frac{4(4+2a)}{(a-2)^2} = 24 \Rightarrow \frac{4+2a}{(a-2)^2} = 6$.$4+2a = 6(a^2-4a+4) \Rightarrow 4+2a = 6a^2-24a+24$.$6a^2-26a+20 = 0 \Rightarrow 3a^2-13a+10 = 0$.Factoring the quadratic: $(3a-10)(a-1) = 0$.So,$a_1=1$ and $a_2=\frac{10}{3}$.Finally,$a_1+3a_2 = 1 + 3(\frac{10}{3}) = 1+10 = 11$.

$ax^2-4xy-2y^2=0$ represents a pair of lines. If $\theta$ is the angle between these lines,$\cos \theta=\frac{1}{5}$ and the possible values of '$a$' are $a_1$ and $a_2$ $(a_1 < a_2)$,then $a_1+3a_2=$

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