Question

Find the locus of intersection of tangents if the difference of these eccentric angles be \[{{120}^{\circ }}\] in ellipse.

Answer 1

Hint: Take two parametric coordinates on the ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] as \[\left( a\cos {{\theta }_{1}},b\sin {{\theta }_{1}} \right)\] and \[\left( a\cos {{\theta }_{2}},b\sin {{\theta }_{2}} \right)\] . Draw tangents through it. Write the equations of tangent by equation T = 0. Solve both of the equations and try to eliminate \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] with the given condition in the problem.

Complete step-by-step answer:
Let us assume ellipse equation as \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1-(1)\left( \because a>b \right)\] .

Let eccentric angles be \[{{\theta }_{1}}\And {{\theta }_{2}}\] and parametric coordinates of A and B are \[\left( a\cos {{\theta }_{1}},b\sin {{\theta }_{1}} \right)\] and \[\left( a\cos {{\theta }_{2}},b\sin {{\theta }_{2}} \right)\] shown in diagram.
As, we know the equation tangent if any point is given on any curve is T=0.
Hence, equations from point A and B as follows: -
 \[\dfrac{a\cos {{\theta }_{1}}x}{{{a}^{2}}}+\dfrac{b\sin {{\theta }_{1}}y}{{{b}^{2}}}=1\]
Or
 \[\dfrac{x\cos {{\theta }_{1}}}{{{a}^{2}}}+\dfrac{b\sin {{\theta }_{1}}}{{{b}^{2}}}=1-(2)\]
Similarly, we can write the equation of second tangent as: -
 \[\dfrac{x\cos {{\theta }_{1}}}{a}+\dfrac{y\sin {{\theta }_{2}}}{b}=1-(3)\]
Now, for getting relation between \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] , we can add both equations (2) and (3) and subtract them as well.
Therefore, adding equation (2) and (3)
 \[\dfrac{x}{a}\left( \cos {{\theta }_{1}}+\cos {{\theta }_{2}} \right)+\dfrac{y}{b}\left( \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right)=2-(4)\]
As we know we have given the relation between \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] is \[{{\theta }_{1}}+{{\theta }_{2}}={{120}^{\circ }}\] ; so we can apply \[\cos x+\cos y\] and \[\sin x+\sin y\] formulae in following way: -
 \[\begin{align}
  & \cos x-\cos y=-2\sin \dfrac{x-y}{2}\sin \dfrac{x+y}{2}-(8) \\
 & \sin x-\sin y=2\sin \dfrac{x-y}{2}\cos \dfrac{x+y}{2}-(8) \\
\end{align}\]
Substituting the values of equations (8) in equation (7) as
 \[\begin{align}
  & \dfrac{x}{a}\left( -2\sin \dfrac{{{\theta }_{1}}-{{\theta }_{2}}}{2}\sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)+\dfrac{y}{b}\left( 2\sin \dfrac{{{\theta }_{1}}-{{\theta }_{2}}}{2}\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)=0 \\
 & \dfrac{x}{a}\sin \left( \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)-\dfrac{y}{b}\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}=0 \\
 & \tan \left( \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)=\dfrac{\dfrac{y}{b}}{\dfrac{x}{a}}=\dfrac{y}{b}\times \dfrac{a}{x}=\dfrac{ay}{bx} \\
\end{align}\]
Now we need to calculate \[\sin \left( \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)\] and \[\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] as we have to put values of it in equation (6)
Let us take a right angled triangle with one acute angle as \[\dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] .

 \[\begin{align}
  & \sin \left( \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)=\dfrac{perpendicular}{Hypo\tan eous}=\dfrac{AB}{AC} \\
 & \cos \left( \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)=\dfrac{Base}{Hypo\tan eous}=\dfrac{BC}{AC} \\
\end{align}\]
We can calculate AC as \[\sqrt{A{{B}^{2}}+B{{C}^{2}}}\] by using Pythagoras theorem.
 \[\begin{align}
  & AC=\sqrt{{{\left( ay \right)}^{2}}+{{\left( bx \right)}^{2}}} \\
 & AC=\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}} \\
\end{align}\]
Hence, we can write \[\sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] and \[\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] as
 \[\begin{align}
  & \sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}=\dfrac{ay}{\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}}-(9) \\
 & \cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}=\dfrac{bx}{\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}}-(9) \\
 & \\
\end{align}\]
Now, we can put values of \[\sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] and \[\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] (calculated in equation (9)) in equation (6).
 \[\begin{align}
  & \cos \dfrac{{{\theta }_{1}}-{{\theta }_{2}}}{2}\left[ \dfrac{x}{a}\cos \cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}+\dfrac{y}{b}\sin \cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right]=1 \\
 & \cos \dfrac{{{\theta }_{1}}-{{\theta }_{2}}}{2}\left[ \dfrac{x}{a}\dfrac{bx}{\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}}+\dfrac{y}{b}\dfrac{ay}{\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}} \right]=1 \\
 & \cos \left( \dfrac{{{\theta }_{1}}-{{\theta }_{2}}}{2} \right)\left[ \dfrac{{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}}{ab\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}} \right]=1 \\
\end{align}\]
Now, we have already given that difference between eccentric angles is 120 i.e. \[{{\theta }_{1}}-{{\theta }_{2}}={{120}^{\circ }}\] .
Hence, we can write the above equation as \[\cos \left( \dfrac{120}{2} \right)\left( \dfrac{\sqrt{{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}}}{ab} \right)=1\]
Squaring both sides and putting the value of \[\cos {{60}^{\circ }}=\dfrac{1}{2}\]
 \[\dfrac{1}{4}\left( {{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}} \right)={{a}^{2}}{{b}^{2}}\]
Or
 \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}=4{{a}^{2}}{{b}^{2}}\] (Required locus)

Note: Another approach for this question would be that we can suppose parametric coordinates \[A(a\cos {{\theta }_{1}},b\sin {{\theta }_{1}})\] and \[B(a\cos {{\theta }_{2}},b\sin {{\theta }_{2}})\] . Now, tangents passing through them can be written by tangent T=0.
Equation of tangents are: -
 \[\begin{align}
  & \dfrac{x\cos {{\theta }_{1}}}{a}+\dfrac{y\sin {{\theta }_{1}}}{b}=1-(1) \\
 & \dfrac{x\cos {{\theta }_{2}}}{a}+\dfrac{y\sin {{\theta }_{2}}}{b}=1-(2) \\
\end{align}\]
Now, find out the intersection of equation (1) and equation (2), then suppose that intersection as h and k and try to get relationship between h and k by elimination of \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] by using the given relationship \[{{\theta }_{1}}-{{\theta }_{2}}={{120}^{\circ }}\] .
As we have to calculate intersection points in the above mentioned method in note but not in actual solution which is a key point of the question and make the solution more flexible.
One can get confuse with the formula of tangent with point given on any curve i.e. T=0.
General way of writing tangent equation if \[\left( {{x}_{1}},{{y}_{1}} \right)\] point lies on curve C then we need to replace
 \[{{x}^{2}}\] by \[x{{x}_{1}}\]
 \[y\] by \[y{{y}_{1}}\]
x by \[\left( \dfrac{x+{{x}_{1}}}{2} \right)\]
y by \[\left( \dfrac{y+{{y}_{1}}}{2} \right)\]
Hence equation of tangent from point \[\left( {{x}_{1}},{{y}_{1}} \right)\] lying on ellipse \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] .