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(A) The pair of lines $A x^2+2 H x y+B y^2=0$ passes through the origin. For these lines to form an equilateral triangle with the line $a x+b y+c=0$,t

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The pair of lines $A x^2+2 H x y+B y^2=0$ passes through the origin. For these lines to form an equilateral triangle with the line $a x+b y+c=0$,the angle between the pair of lines must be $60^\circ$ or $\frac{\pi}{3}$ radians.The angle $	heta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $	an 	heta = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.Setting $	heta = \frac{\pi}{3}$,we have $	an \frac{\pi}{3} = \sqrt{3}$.So,$\sqrt{3} = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$.$3(A+B)^2 = 4(H^2-A B)$.$3(A^2+2 A B+B^2) = 4 H^2-4 A B$.$3 A^2+6 A B+3 B^2 = 4 H^2-4 A B$.$3 A^2+10 A B+3 B^2 = 4 H^2$.Factoring the left side: $3 A^2+9 A B+A B+3 B^2 = 4 H^2$.$3 A(A+3 B)+B(A+3 B) = 4 H^2$.$(A+3 B)(3 A+B) = 4 H^2$.Thus,the correct option is $A$.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

If the pair of straight lines given by $A x^2+2 H x y+B y^2=0$,where $(H^2>A B)$,forms an equilateral triangle with the line $a x+b y+c=0$,then $(A+3 B)(3 A+B)=$ (in $H^2$)

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