
If the angle \[2\theta \] is acute, then the acute angle between ${x^2}\left( {\cos \theta - \sin \theta } \right) + 2xy\cos \theta + {y^2}\left( {\cos \theta + \sin \theta } \right) = 0$ is
A. \[2\theta \]
B. $\dfrac{\theta }{3}$
C. $\theta $
D. $\dfrac{\theta }{2}$
Answer
162k+ views
Hint: In order to solve this type of question, first we need to write the given equation and the general equation. Then, we have to compare both the equations to get the values of a, h and b. Next, we will use the formula for finding the angle between the lines and substitute the values in it to get the correct answer.
Formula used:
$a{x^2} + 2hxy + b{y^2} = 0$
$\tan \varphi = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution:
We are given the following pair of lines,
${x^2}\left( {\cos \theta - \sin \theta } \right) + 2xy\cos \theta + {y^2}\left( {\cos \theta + \sin \theta } \right) = 0$ ………………….equation$\left( 1 \right)$
We know that the general equation is,
$a{x^2} + 2hxy + b{y^2} = 0$ ………………….equation$\left( 2 \right)$
On comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = \left( {\cos \theta - \sin \theta } \right),\;b = \left( {\cos \theta + \sin \theta } \right),\;h = \cos \theta $
Finding the angle between the lines,
$\tan \varphi = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Substituting the values of a, h and b,
\[\tan \varphi = \left| {\dfrac{{2\sqrt {{{\left( {\cos \theta } \right)}^2} - \left( {\cos \theta - \sin \theta } \right) \times \left( {\cos \theta + \sin \theta } \right)} }}{{\left( {\cos \theta - \sin \theta } \right) + \left( {\cos \theta + \sin \theta } \right)}}} \right|\]
$\tan \varphi = \left| {\dfrac{{2\sqrt {{{\cos }^2}\theta - \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)} }}{{2\cos \theta }}} \right|$
Solving it,
$\tan \varphi = \left| {\dfrac{{\sqrt {{{\sin }^2}\theta } }}{{\cos \theta }}} \right|$
$\tan \varphi = \left| {\dfrac{{\sin \theta }}{{\cos \theta }}} \right|$
Simplifying it further,
$\tan \varphi = \tan \theta $
On comparing we get,
$\varphi = \theta $
$\therefore $ The correct option is C.
Note: Avoid any type of calculation mistakes while substituting and solving the equation to get the correct angle. Also, one has to be careful while writing the general equation and comparing the values of a, h and b.
Formula used:
$a{x^2} + 2hxy + b{y^2} = 0$
$\tan \varphi = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution:
We are given the following pair of lines,
${x^2}\left( {\cos \theta - \sin \theta } \right) + 2xy\cos \theta + {y^2}\left( {\cos \theta + \sin \theta } \right) = 0$ ………………….equation$\left( 1 \right)$
We know that the general equation is,
$a{x^2} + 2hxy + b{y^2} = 0$ ………………….equation$\left( 2 \right)$
On comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$a = \left( {\cos \theta - \sin \theta } \right),\;b = \left( {\cos \theta + \sin \theta } \right),\;h = \cos \theta $
Finding the angle between the lines,
$\tan \varphi = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Substituting the values of a, h and b,
\[\tan \varphi = \left| {\dfrac{{2\sqrt {{{\left( {\cos \theta } \right)}^2} - \left( {\cos \theta - \sin \theta } \right) \times \left( {\cos \theta + \sin \theta } \right)} }}{{\left( {\cos \theta - \sin \theta } \right) + \left( {\cos \theta + \sin \theta } \right)}}} \right|\]
$\tan \varphi = \left| {\dfrac{{2\sqrt {{{\cos }^2}\theta - \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)} }}{{2\cos \theta }}} \right|$
Solving it,
$\tan \varphi = \left| {\dfrac{{\sqrt {{{\sin }^2}\theta } }}{{\cos \theta }}} \right|$
$\tan \varphi = \left| {\dfrac{{\sin \theta }}{{\cos \theta }}} \right|$
Simplifying it further,
$\tan \varphi = \tan \theta $
On comparing we get,
$\varphi = \theta $
$\therefore $ The correct option is C.
Note: Avoid any type of calculation mistakes while substituting and solving the equation to get the correct angle. Also, one has to be careful while writing the general equation and comparing the values of a, h and b.
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