
If \[\theta \] is the angle between the curves \[xy = 2\] and \[{x^2} + 4y = 0\] then \[\tan \theta
\] is equal to
1.\[1\]
2.\[ - 1\]
3.\[2\]
4.\[3\]
Answer
435.9k+ views
Hint: Firstly find the point of intersection of the given curves. Differentiate both the curves and find the value of the derivative at the point of intersection. You will get the respective slopes of the lines. Then apply the \[\tan \theta \] formula related to the slopes.
Complete step-by-step solution:
Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between two distinct points on a line
A line is increasing if it goes up from left to right. The slope is positive, i.e. \[m > 0\]
A line is decreasing if it goes down from left to right. The slope is negative, i.e. \[m < 0\] .
If a line is horizontal the slope is zero. This is a constant function.
If a line is vertical the slope is undefined.
Finding the points of intersection of both the curves \[xy = 2\] and \[{x^2} + 4y = 0\]
\[x = - 2\] and \[y = - 1\]
Hence we get the point of intersection of the given curves as \[( - 2, - 1)\].
Consider the line \[xy = 2\]
We have \[y = \dfrac{2}{x}\]
Differentiating both the sides with respect to \[x\]we get ,
\[\dfrac{{dy}}{{dx}} = - \dfrac{2}{{{x^2}}}\]
\[\Rightarrow {\dfrac{{dy}}{{dx}}_{(x = - 2)}} = - \dfrac{2}{4} = - \dfrac{1}{2}\]
Consider the line\[{x^2} + 4y = 0\]
We have \[y = - \dfrac{{{x^2}}}{4}\]
Differentiating both the sides with respect to \[x\] we get ,
\[\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{4} = - \dfrac{x}{2}\]
\[\Rightarrow {\dfrac{{dy}}{{dx}}_{(x = - 2)}} = 1\]
Therefore we get \[{m_1} = - \dfrac{1}{2}\] and \[{m_2} = 1\]
We know that \[\tan \theta = \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}\] where \[\theta \] is the angle between the given curves.
\[\tan \theta = \dfrac{{1 - \left( { - \dfrac{1}{2}} \right)}}{{1 + (1)\left( { - \dfrac{1}{2}} \right)}} = \dfrac{{\dfrac{3}{2}}}{{\dfrac{1}{2}}} = 3\]
Therefore option (4) is the correct answer.
Note: We must have a strong grip over the concepts of trigonometry, related formulae to do these types of questions. we should also know the derivatives of certain functions correctly.we should take care of calculations while solving such questions.
Complete step-by-step solution:
Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between two distinct points on a line
A line is increasing if it goes up from left to right. The slope is positive, i.e. \[m > 0\]
A line is decreasing if it goes down from left to right. The slope is negative, i.e. \[m < 0\] .
If a line is horizontal the slope is zero. This is a constant function.
If a line is vertical the slope is undefined.
Finding the points of intersection of both the curves \[xy = 2\] and \[{x^2} + 4y = 0\]
\[x = - 2\] and \[y = - 1\]
Hence we get the point of intersection of the given curves as \[( - 2, - 1)\].
Consider the line \[xy = 2\]
We have \[y = \dfrac{2}{x}\]
Differentiating both the sides with respect to \[x\]we get ,
\[\dfrac{{dy}}{{dx}} = - \dfrac{2}{{{x^2}}}\]
\[\Rightarrow {\dfrac{{dy}}{{dx}}_{(x = - 2)}} = - \dfrac{2}{4} = - \dfrac{1}{2}\]
Consider the line\[{x^2} + 4y = 0\]
We have \[y = - \dfrac{{{x^2}}}{4}\]
Differentiating both the sides with respect to \[x\] we get ,
\[\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{4} = - \dfrac{x}{2}\]
\[\Rightarrow {\dfrac{{dy}}{{dx}}_{(x = - 2)}} = 1\]
Therefore we get \[{m_1} = - \dfrac{1}{2}\] and \[{m_2} = 1\]
We know that \[\tan \theta = \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}\] where \[\theta \] is the angle between the given curves.
\[\tan \theta = \dfrac{{1 - \left( { - \dfrac{1}{2}} \right)}}{{1 + (1)\left( { - \dfrac{1}{2}} \right)}} = \dfrac{{\dfrac{3}{2}}}{{\dfrac{1}{2}}} = 3\]
Therefore option (4) is the correct answer.
Note: We must have a strong grip over the concepts of trigonometry, related formulae to do these types of questions. we should also know the derivatives of certain functions correctly.we should take care of calculations while solving such questions.
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