Question

Find the angle between the lines joining the points of intersection of line \[y = 3x + 2\] and the curve \[{x^2} + 2xy + 3{y^2} + 4x + 8y - 11 = 0\] to the origin.
A. \[\tan^{ - 1}\left( {\dfrac{3}{{2\sqrt 2 }}} \right)\]
B. \[\tan^{ - 1}\left( {\dfrac{{2\sqrt 2 }}{3}} \right)\]
C. \[\tan^{ - 1}\left( {\sqrt 3 } \right)\]
D. \[\tan^{ - 1}\left( {2\sqrt 2 } \right)\]

Answer 1

Hint In the given question, the equation of a straight line and the equation of the curve are given. Convert the equation of the curve into a homogeneous equation of second degree in \[x\] and \[y\] by using the equation of a straight line. Then compare the homogeneous equation with the general homogeneous equation. By using the formula of the angle between the pair of lines, find the angle between the lines joining the points of intersection of the given lines and origin.

Formula used:
General homogeneous equation: \[a{x^2} + 2hxy + b{y^2} = 0\]
Angle between the pair of lines: \[\theta = \tan^{ - 1}\left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|\]

Complete step by step solution:
The given equation of a straight line is \[y = 3x + 2\] and the equation of the curve is \[{x^2} + 2xy + 3{y^2} + 4x + 8y - 11 = 0\].
Let’s simplify the equation of straight line.
\[y = 3x + 2\]
\[ \Rightarrow \]\[\dfrac{{y - 3x}}{2} = 1\]

Now convert the equation of curve into a homogeneous equation of second degree in \[x\] and \[y\] by using the above equation.
\[{x^2} + 2xy + 3{y^2} + 4x\left( 1 \right) + 8y\left( 1 \right) - 11{\left( 1 \right)^2} = 0\]
\[ \Rightarrow \]\[{x^2} + 2xy + 3{y^2} + 4x\left( {\dfrac{{y - 3x}}{2}} \right) + 8y\left( {\dfrac{{y - 3x}}{2}} \right) - 11{\left( {\dfrac{{y - 3x}}{2}} \right)^2} = 0\]
\[ \Rightarrow \]\[ - 5{x^2} + 7{y^2} - 8xy - 11\left( {\dfrac{{{y^2} - 6xy + 9{x^2}}}{4}} \right) = 0\]
Multiply both sides by 4.
\[ - 20{x^2} + 28{y^2} - 32xy - 11\left( {{y^2} - 6xy + 9{x^2}} \right) = 0\]
\[ \Rightarrow \]\[ - 20{x^2} + 28{y^2} - 32xy - 11{y^2} + 66xy - 99{x^2} = 0\]
\[ \Rightarrow \]\[ - 119{x^2} + 17{y^2} + 34xy = 0\]
\[ \Rightarrow \]\[119{x^2} - 17{y^2} - 34xy = 0\]
Divide both sides by \[17\].
\[7{x^2} - {y^2} - 2xy = 0\]

Compare the above equation with the general equation \[a{x^2} + 2hxy + b{y^2} = 0\].
Then,
\[a = 7, b = - 1, h = - 1\]

Now apply the formula \[\theta = ta{n^{ - 1}}\left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|\] to calculate the angle between the pair of lines.
\[\theta = \tan^{ - 1}\left| {\dfrac{{2\sqrt {{{\left( { - 1} \right)}^2} - \left( 7 \right)\left( { - 1} \right)} }}{{7 + \left( { - 1} \right)}}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{{2\sqrt {1 + 7} }}{6}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{{\sqrt 8 }}{3}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{{2\sqrt 2 }}{3}} \right|\]

Hence the correct option is option B.

Note: The angle between the two lines can be found by calculating the slope of each line and then using them in the formula to determine the angle between two lines when the slope of each line is known.