If the pair of straight lines given by A x^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 $A x^2+2 H x y+B y^2=0$. Since the triangle formed by these lines and $a x+b y+c=0$ is equilateral,the angle between

Vedclass · Accepted Answer

$4 H^2$

Vedclass · Answer

(D) The pair of lines is given by $A x^2+2 H x y+B y^2=0$. Since the triangle formed by these lines and $a x+b y+c=0$ is equilateral,the angle between the lines of the pair must be $60^{\circ}$. 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|}$. For an equilateral triangle,$	heta = 60^{\circ}$,so $	an 60^{\circ} = \sqrt{3} = \frac{2\sqrt{H^2-A B}}{|A+B|}$. Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$. $3(A+B)^2 = 4(H^2-A B)$. $3(A^2+B^2+2 A B) = 4 H^2-4 A B$. $3 A^2+3 B^2+6 A B = 4 H^2-4 A B$. $3 A^2+10 A B+3 B^2 = 4 H^2$. Factoring the left side,we get $(3 A+B)(A+3 B) = 4 H^2$.

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

Similar Questions

What is the scientific name of modern man?

The quantum number which explains the line spectra observed as doublets in the case of hydrogen and alkali metals and doublets and triplets in the case of alkaline earth metals is

The time in seconds required to produce a potential difference of $20 \ V$ across a capacitor of $1000 \ \mu F$ when it is charged at a steady rate of $200 \ \mu C/s$ is:

Reduction in $pH$ of blood will

Columella auris is a modified

Vedclass Test Series

Exam Paper Generator

Online Exam Module