Question

The angle between ellipse \[\dfrac{{{x^2}}}{4} + {y^2} = 1\] and circle \[{x^2} + {y^2} = 2\] is \[\theta \] , then \[\theta \] is equal to
A \[\dfrac{1}{2}\]
B \[\dfrac{1}{{\sqrt 2 }}\]
C \[\dfrac{1}{{2\sqrt 2 }}\]

Answer 1

Hint: To find the angle \[\theta \] , we must find the point of intersection for both the curves, then the equation of tangent and its slope. Hence, with respect to the slope we need to find the value of \[\theta \] by applying \[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\] , we need to substitute the values of slope.
Formula used:
 \[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\]
 \[\theta \] is the angle between ellipse and circle.
 \[{m_1}{m_2}\] are the slopes.

Complete step-by-step answer:
Given,
 \[\dfrac{{{x^2}}}{4} + {y^2} = 1\]
 \[ \to \] \[\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\] ………… 1
 \[ \Rightarrow \] \[{x^2} + {y^2} = 2\]
Hence, \[{y^2} = 2 - {x^2}\] …………… 2
Now, substitute the value of \[{y^2}\] in equation 1 i.e.,
 \[\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\]
 \[ \Rightarrow \] \[\dfrac{{{x^2}}}{4} + \dfrac{{2 - {x^2}}}{1} = 1\]
Simplify the terms:
 \[{x^2} + 8 - 4{x^2} = 4\]
 \[ - 3{x^2} = 4 - 8\]
 \[ - 3{x^2} = - 4\]
Now, divide both sides of the expression by -3, as
 \[\dfrac{{ - 3{x^2}}}{{ - 3}} = \dfrac{{ - 4}}{{ - 3}}\]
We can see that numerator and denominator terms are same, hence -3 and -3 implies to 1, we get
 \[{x^2} = \dfrac{{ - 4}}{{ - 3}}\]
 \[ \Rightarrow {x^2} = \dfrac{4}{3}\]
Therefore, the value of x is:
 \[ \Rightarrow x = \pm \dfrac{2}{{\sqrt 3 }}\]
Now, to obtain the value of y, substitute the value of x in equation 2 as:
 \[{y^2} = 2 - {x^2}\]
 \[{y^2} = 2 - {\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2}\]
Simplifying the terms, we get:
 \[{y^2} = 2 - \dfrac{4}{3}\]
 \[{y^2} = \dfrac{2}{3}\]
Therefore, the value of y is:
 \[y = \pm \sqrt {\dfrac{2}{3}} \]
The point of intersection for both the curves are:
 \[\left( {\dfrac{2}{{\sqrt 3 }},\sqrt {\dfrac{2}{3}} } \right)\] , \[\left( {\dfrac{2}{{\sqrt 3 }}, - \sqrt {\dfrac{2}{3}} } \right)\] , \[\left( {\dfrac{{ - 2}}{{\sqrt 2 }},\sqrt {\dfrac{2}{3}} } \right)\] , \[\left( {\dfrac{{ - 2}}{{\sqrt 3 }}, - \sqrt {\dfrac{2}{3}} } \right)\]
Let,
 \[P\left( {\dfrac{2}{{\sqrt 3 }},\sqrt {\dfrac{2}{3}} } \right)\] at the ellipse \[\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{1} = 1\] .
Then the equation of tangent is:
 \[\dfrac{x}{4}\left( {\dfrac{2}{{\sqrt 3 }}} \right) + y\left( {\sqrt {\dfrac{2}{3}} } \right) = 1\]
 \[ \Rightarrow \] \[\dfrac{x}{{2\sqrt 3 }} + \dfrac{{\sqrt 2 y}}{{\sqrt 3 }} = 1\] …………….. 3
Let,
 \[P\left( {\dfrac{2}{{\sqrt 3 }},\sqrt {\dfrac{2}{3}} } \right)\] at the circle \[{x^2} + {y^2} = 2\] .
Then the equation of tangent is:
 \[x\left( {\dfrac{2}{{\sqrt 3 }}} \right) + y\left( {\sqrt {\dfrac{2}{3}} } \right) = 2\]
 \[ \Rightarrow \] \[\dfrac{2}{{\sqrt 3 }}x + \dfrac{{\sqrt 2 }}{{\sqrt 3 }}y = 2\] ………….. 4
Hence, the slope of equation 3, with respect to x and y is:
 \[\dfrac{x}{{2\sqrt 3 }} + \dfrac{{\sqrt 2 y}}{{\sqrt 3 }} = 1\]
 \[ \Rightarrow \] \[{m_1} = \dfrac{{ - \dfrac{1}{{2\sqrt 3 }}}}{{\dfrac{{\sqrt 2 }}{{\sqrt 3 }}}}\]
 \[ \Rightarrow \] \[{m_1} = \dfrac{{ - 1}}{{2\sqrt 2 }}\]
And, the slope of equation 4, with respect to x and y is:
 \[\dfrac{2}{{\sqrt 3 }}x + \dfrac{{\sqrt 2 }}{{\sqrt 3 }}y = 2\]
 \[ \Rightarrow \] \[{m_2} = \dfrac{{ - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 2 }}{{\sqrt 3 }}}}\]
As \[\sqrt 3 \] is a common term in both numerator and denominator which implies to one, hence we get
 \[ \Rightarrow \] \[{m_2} = \dfrac{{ - 2}}{{\sqrt 2 }}\]
 \[ \Rightarrow \] \[{m_2} = - \sqrt 2 \]
Now, with respect to the slope we must find the angle between ellipse and circle is given as:
 \[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\]
Now, substitute the value of slope m1 and m2 we get:
 \[\tan \theta = \left| {\dfrac{{\dfrac{{ - 1}}{{2\sqrt 2 }} - \left( { - \sqrt 2 } \right)}}{{1 + \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right)\left( { - \sqrt 2 } \right)}}} \right|\]
 \[\tan \theta = \left| {\dfrac{{\dfrac{{ - 1}}{{2\sqrt 2 }} + \sqrt 2 }}{{1 + \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right)\left( { - \sqrt 2 } \right)}}} \right|\]
Simplify the terms, of the expression we get:
 \[\tan \theta = \left| {\dfrac{{\dfrac{{ - 1 + 4}}{{2\sqrt 2 }}}}{{\dfrac{{2\sqrt 2 + \sqrt 2 }}{{2\sqrt 2 }}}}} \right|\]
 \[\tan \theta = \left| {\dfrac{3}{{3\sqrt 2 }}} \right|\]
 \[\tan \theta = \left| {\dfrac{1}{{\sqrt 2 }}} \right|\]
Hence, we get:
 \[\tan \theta = \dfrac{1}{{\sqrt 2 }}\]
So, the correct answer is “Option B”.

Note: The key point to solve this question is that we must know all the basic relations of the ellipse with respect to the circle and its slope form to find the angle we need slope. We must also note the point of intersection for both the curves i.e., points at the ellipse and circle to find the equation of tangent. If we are asked to find the angle of the circle and if we know the cartesian coordinates of the point x, y then you can find the angle theta by converting it into polar coordinates.