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

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

Question

(D) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $ax + by + c = 0$ if and only if the angle between the line

Vedclass · Accepted Answer

$4H^2$

Vedclass · Answer

(D) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $ax + by + c = 0$ if and only if the angle between the lines is $60^\circ$ and the lines are symmetric with respect to the line $ax + by + c = 0$.This condition is satisfied when the pair of lines is given by $(a^2 - 3b^2)x^2 + 8abxy + (b^2 - 3a^2)y^2 = 0$.Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we get:$A = a^2 - 3b^2$,$B = b^2 - 3a^2$,and $2H = 8ab$.Now,calculate $(A + 3B)(3A + B)$:$A + 3B = (a^2 - 3b^2) + 3(b^2 - 3a^2) = a^2 - 3b^2 + 3b^2 - 9a^2 = -8a^2$.$3A + B = 3(a^2 - 3b^2) + (b^2 - 3a^2) = 3a^2 - 9b^2 + b^2 - 3a^2 = -8b^2$.Therefore,$(A + 3B)(3A + B) = (-8a^2)(-8b^2) = 64a^2b^2$.Since $2H = 8ab$,we have $H = 4ab$,so $H^2 = 16a^2b^2$.Thus,$64a^2b^2 = 4(16a^2b^2) = 4H^2$.

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

Which of the following effects is observed after the consumption of $\Delta-9-THC$?

The value of $\lambda$ such that $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$ represents a pair of straight lines is:

An ideal gas undergoes a thermodynamic cycle as shown in the figure. Which of the following graphs represents the same cycle?

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane, where $a, b, c$ $(a < b < c)$ are the factors of $2001$, then the equation of the plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module