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 $Ax^2+2Hxy+By^2=0$. Since the triangle formed by these lines and $ax+by+c=0$ is equilateral,the angle between the li

Vedclass · Accepted Answer

$4H^2$

Vedclass · Answer

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

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 pair of substances gives the same gaseous product when they react with water?

Which one of the following statements is not correct?

If $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$,then:

$3.2$ moles of hydrogen iodide were heated in a sealed bulb at $444 \, ^oC$ until the equilibrium state was reached. Its degree of dissociation at this temperature was found to be $22 \%$. The number of moles of hydrogen iodide present at equilibrium are:

If $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$,then the angle between $x^2+2 x y \sec \theta+y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module