Question

Find the condition of collinearity of the points $\left( a\cos {{\phi }_{1}},b\sin {{\phi }_{1}} \right),\left( a\cos {{\phi }_{2}},b\sin {{\phi }_{2}} \right),\left( a\cos {{\phi }_{3}},b\sin {{\phi }_{3}} \right)$ \[\]

Answer 1

Hint: We denote the three points as $A\left( a\cos {{\phi }_{1}},b\sin {{\phi }_{1}} \right),B\left( a\cos {{\phi }_{2}},b\sin {{\phi }_{2}} \right)$ and $C\left( a\cos {{\phi }_{3}},b\sin {{\phi }_{3}} \right)$. We denote the slopes of AB, BC, CA as ${{m}_{1}},{{m}_{2}},{{m}_{3}}$. We use the fact that three points will lie on a line when the all the slopes of lines joining any two point on the main line will be equal, We solve ${{m}_{1}}={{m}_{2}}={{m}_{3}}$ to get the condition of collinearity. \[\]

Complete step by step answer:
Let us denote the three points as $A\left( a\cos {{\phi }_{1}},b\sin {{\phi }_{1}} \right),B\left( a\cos {{\phi }_{2}},b\sin {{\phi }_{2}} \right)$ and $C\left( a\cos {{\phi }_{3}},b\sin {{\phi }_{3}} \right)$. If they are collinear then the three points $A,B,C$ will lie in the same line which means the slope of the line which contains $A,B,C$ will be equal to the slope of $AB,BC,AC$.\[\]
We know that the slope of line joining two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$is given by
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Let us denote the slope of AB as ${{m}_{1}}$, slope of BC as ${{m}_{2}}$and slope of CA ${{m}_{3}}$. We have,
\[{{m}_{1}}=\dfrac{b\left( \sin {{\phi }_{2}}-\sin {{\phi }_{1}} \right)}{a\left( \cos {{\phi }_{2}}-\cos {{\phi }_{1}} \right)}=\dfrac{b\cos \left( \dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)\sin \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)}{-a\sin \left( \dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)\sin \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)}=\dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)\]
We can similarly find the slopes of BC and CA as
\[\begin{align}
  & {{m}_{2}}=\dfrac{b\left( \sin {{\phi }_{2}}-\sin {{\phi }_{3}} \right)}{a\left( \cos {{\phi }_{2}}-\cos {{\phi }_{3}} \right)}=\dfrac{b\cos \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right)\sin \left( \dfrac{{{\phi }_{2}}-{{\phi }_{3}}}{2} \right)}{-a\sin \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right)\sin \left( \dfrac{{{\phi }_{2}}-{{\phi }_{3}}}{2} \right)}=\dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right) \\
 & {{m}_{3}}=\dfrac{b\left( \sin {{\phi }_{1}}-\sin {{\phi }_{3}} \right)}{a\left( \cos {{\phi }_{1}}-\cos {{\phi }_{3}} \right)}=\dfrac{b\cos \left( \dfrac{{{\phi }_{1}}+{{\phi }_{3}}}{2} \right)\sin \left( \dfrac{{{\phi }_{1}}-{{\phi }_{3}}}{2} \right)}{-a\sin \left( \dfrac{{{\phi }_{1}}+{{\phi }_{3}}}{2} \right)\sin \left( \dfrac{{{\phi }_{1}}-{{\phi }_{3}}}{2} \right)}=\dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{1}}+{{\phi }_{3}}}{2} \right) \\
\end{align}\]
We have equality of slopes from the condition of collinearity which is ${{m}_{1}}={{m}_{2}}={{m}_{3}}$. We take ${{m}_{1}}={{m}_{2}}$ and use the relation between tangent and cotangent of an angle to have ,
\[\begin{align}
  & \dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)=\dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right) \\
 & \Rightarrow \tan \left( \dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)=\tan \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right)\left( \tan \theta =\dfrac{1}{\cot \theta } \right) \\
 & \Rightarrow {{\phi }_{1}}+{{\phi }_{2}}={{\phi }_{2}}+{{\phi }_{3}} \\
 & \Rightarrow {{\phi }_{1}}={{\phi }_{3}} \\
\end{align}\]
We can similarly take ${{m}_{2}}={{m}_{3}}$ to have
\[\begin{align}
  & \dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right)=\dfrac{-b}{a}\cot \left( \dfrac{{{\phi }_{1}}+{{\phi }_{3}}}{2} \right) \\
 & \Rightarrow \tan \left( \dfrac{{{\phi }_{2}}+{{\phi }_{3}}}{2} \right)=\tan \left( \dfrac{{{\phi }_{1}}+{{\phi }_{3}}}{2} \right)\left( \tan \theta =\dfrac{1}{\cot \theta } \right) \\
 & \Rightarrow {{\phi }_{2}}+{{\phi }_{3}}={{\phi }_{1}}+{{\phi }_{3}} \\
 & \Rightarrow {{\phi }_{2}}={{\phi }_{1}} \\
\end{align}\]
So the condition of collinearity is ${{\phi }_{1}}={{\phi }_{2}}={{\phi }_{3}}$. If we take ${{\phi }_{1}}={{\phi }_{2}}={{\phi }_{3}}=\phi $ the coordinates will be $A\left( a\cos \phi ,b\sin \phi \right),B\left( a\cos \phi ,b\sin \phi \right),C \left( a\cos \phi ,b\sin \phi \right)$.So they are identical.\[\]

Note: We note that all the solutions of the equation $\tan x=\tan a$ are $x=n\pi +a$ where $n$ is an integer, we have rejected the rest of the solution to find the required condition. We note that the given points are in the form of $\left( a\cos t,b\sin t \right)$ with $t$ as a parameter is general coordinate of a point lying on an ellipse whose semi-major axis has length $a$ and semi-minor axis has length $b$ . We can alternatively find the co-linearity condition by equating the area of a triangle with three points $\left( {{x}_{1}},{{y}_{2}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ to zero which means $A=\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)=0$