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 the angle $2 \theta$ is acute,then the acute angle between the pair of straight lines $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$ is:

If the pair of straight lines $6x^2 - 5xy + y^2 = 0$ makes angles $\alpha$ and $\beta$ with the $X$-axis,then $\tan(\alpha - \beta) = $

Which of the following equations represents a pair of straight lines perpendicular to each other?

If the lines represented by $(1+\sin^2 \theta) x^2+2hxy+2\sin \theta y^2=0$,where $\theta \in [0, 2\pi]$,are perpendicular to each other,then $\theta = \dots$.

The angle between the pair of straight lines $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module