If the acute angles between the pairs of line | vedclass

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

Question

(A) The acute angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} igh

Vedclass · Accepted Answer

$	heta_1 = 	heta_2$

Vedclass · Answer

(A) The acute angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.For the first pair $3x^2 - 7xy + 4y^2 = 0$,we have $a=3, 2h=-7, b=4$. Thus $h = -3.5$.$	an 	heta_1 = \left| \frac{2\sqrt{(-3.5)^2 - (3)(4)}}{3 + 4} ight| = \left| \frac{2\sqrt{12.25 - 12}}{7} ight| = \frac{2\sqrt{0.25}}{7} = \frac{2(0.5)}{7} = \frac{1}{7}$.So,$	heta_1 = 	an^{-1}(\frac{1}{7})$.For the second pair $6x^2 - 5xy + y^2 = 0$,we have $a=6, 2h=-5, b=1$. Thus $h = -2.5$.$	an 	heta_2 = \left| \frac{2\sqrt{(-2.5)^2 - (6)(1)}}{6 + 1} ight| = \left| \frac{2\sqrt{6.25 - 6}}{7} ight| = \frac{2\sqrt{0.25}}{7} = \frac{2(0.5)}{7} = \frac{1}{7}$.So,$	heta_2 = 	an^{-1}(\frac{1}{7})$.Therefore,$	heta_1 = 	heta_2$.

If the acute angles between the pairs of lines $3x^2 - 7xy + 4y^2 = 0$ and $6x^2 - 5xy + y^2 = 0$ are $\theta_1$ and $\theta_2$ respectively,then:

Similar Questions

If $\theta$ is the angle between the two lines represented by $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,then what is the value of $\text{cosec}^2\theta$? (where $\lambda$ is a real number.)

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

The angle between the lines represented by the equation $4x^2 - 24xy + 11y^2 = 0$ is given by:

If the slope of one of the lines represented by $2x^2 + 3xy + ky^2 = 0$ is $2$,then the angle between the pair of lines is

If $(m+3n)(3m+n)=4h^2$,then the acute angle between the lines represented by $mx^2+2hxy+ny^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module