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

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively,then $P(X \geq 1)$ is equal to

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

If $\tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3$,then $\tan 3x$ is equal to

If $q$ is false and $p \wedge q \leftrightarrow r$ is true,then which one of the following statements is a tautology?

Vedclass Test Series

Exam Paper Generator

Online Exam Module