(A) The given equation is $x^2 + 2xy \sec \theta + y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$h = \sec

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(A) The given equation is $x^2 + 2xy \sec \theta + y^2 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$h = \sec \theta$,and $b = 1$.Let $\alpha$ be the angle between the lines. The formula for the angle is $\tan \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.Substituting the values: $\tan \alpha = \left| \frac{2\sqrt{\sec^2 \theta - 1}}{1 + 1} \right|$.Since $\sec^2 \theta - 1 = \tan^2 \theta$,we have $\tan \alpha = \left| \frac{2\sqrt{\tan^2 \theta}}{2} \right| = |\tan \theta|$.Therefore,$\alpha = \theta$.

Class 11 Mathematics Pair of straight lines Angle between the pair of straight lines, Condition for parallel and perpendicular lines

Medium

The angle between the lines represented by the equation $x^2 + 2xy \sec \theta + y^2 = 0$ is

A
$\theta$
B
$2\theta$
C
$\frac{\theta}{2}$
D
None of these

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Class 11

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Pair of straight lines

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Angle between the pair of straight lines, Condition for parallel and perpendicular lines

The angle $\theta$ between the pair of straight lines represented by the homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is given by:

Medium

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Four different pairs of lines are given in List-$I$ and the cosine of the angle between every pair of lines is given in List-$II$. Match the following:
List-$I$ List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$ $(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$ $(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$ $(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$ $(IV)$ $\frac{2}{3}$
$(V)$ $\frac{1}{\sqrt{2}}$

The correct match is:

EasyTS EAMCET 2022

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If the pair of lines $2x^2 + 3xy + y^2 = 0$ makes angles $\theta_1$ and $\theta_2$ with the positive direction of the $X$-axis,then $|\tan(\theta_1 - \theta_2)| = $

MediumAP EAMCET 2017

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If the pair of lines given by $(x^2+y^2) \cos^2 \theta = (x \cos \theta + y \sin \theta)^2$ are perpendicular to each other,then $\theta$ is equal to

DifficultAP EAMCET 2012

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Find the angle between the lines represented by $2x^2 - 7xy + 3y^2 = 0$ in degrees. (in $^o$)

Easy

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List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

The angle between the lines represented by the equation $x^2 + 2xy \sec \theta + y^2 = 0$ is

Similar Questions

The angle $\theta$ between the pair of straight lines represented by the homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is given by:

If the pair of lines $2x^2 + 3xy + y^2 = 0$ makes angles $\theta_1$ and $\theta_2$ with the positive direction of the $X$-axis,then $|\tan(\theta_1 - \theta_2)| = $

If the pair of lines given by $(x^2+y^2) \cos^2 \theta = (x \cos \theta + y \sin \theta)^2$ are perpendicular to each other,then $\theta$ is equal to

Find the angle between the lines represented by $2x^2 - 7xy + 3y^2 = 0$ in degrees. (in $^o$)

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