If the pair of straight lines given by Ax^2+2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Let the angle between these lines be $2	heta$. Since the triangle formed with the line $ax+by+c

Vedclass · Accepted Answer

$4H^2$

Vedclass · Answer

(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$. Let the angle between these lines be $2	heta$. Since the triangle formed with the line $ax+by+c=0$ is equilateral,the angle between each pair of lines is $60^{\circ}$. The angle $	heta$ between the lines $y=m_1x$ and $y=m_2x$ is given by $	an 	heta = \left|\frac{2\sqrt{H^2-AB}}{A+B}ight|$. For an equilateral triangle,the angle between the lines is $60^{\circ}$,so $	heta = 30^{\circ}$. Thus,$	an 30^{\circ} = \frac{\sqrt{H^2-AB}}{|A+B|}$. Squaring both sides,we get $\frac{1}{3} = \frac{H^2-AB}{(A+B)^2}$. $(A+B)^2 = 3(H^2-AB)$. $A^2+B^2+2AB = 3H^2-3AB$. $A^2+B^2+5AB = 3H^2$. We need to evaluate $(A+3B)(3A+B) = 3A^2+AB+9AB+3B^2 = 3A^2+10AB+3B^2$. From the condition of the angle between the lines $Ax^2+2Hxy+By^2=0$ being $60^{\circ}$,we use the formula $\cos 60^{\circ} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}}$. $\frac{1}{2} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}} \implies (A-B)^2+4H^2 = 4(A+B)^2$. $A^2+B^2-2AB+4H^2 = 4(A^2+B^2+2AB)$. $4H^2 = 3A^2+3B^2+10AB$. Therefore,$(A+3B)(3A+B) = 3A^2+10AB+3B^2 = 4H^2$.

If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2>AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

Similar Questions

If the lines represented by $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ intersect on the $x$-axis,which of the following is in general incorrect?

Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is

The line $x+y=1$ meets the lines represented by the equation $y^3-6xy^2+11x^2y-6x^3=0$ at the points $P, Q, R$. If $O$ is the origin,then $(OP)^2+(OQ)^2+(OR)^2=$

If the lines represented by $2x^2 + 6xy + y^2 = 0$ are $L_1$ and $L_2$,and the lines represented by $4x^2 + 18xy + y^2 = 0$ are $l_1$ and $l_2$,and the acute angle between $L_1$ and $l_1$ is $\theta$,then the acute angle between $L_2$ and $l_2$ is :-

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

Vedclass Test Series

Exam Paper Generator

Online Exam Module