Question

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

Answer 1

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: -

& \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{aligned}$$ Substituting the values of equations (8) in equation (7) as $$\begin{aligned} & \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{aligned}$$ Now we need to calculate $$\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{aligned} & \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{aligned}$$ We can calculate AC as $$\sqrt{A{{B}^{2}}+B{{C}^{2}}}$$ by using Pythagoras theorem. $$\begin{aligned} & AC=\sqrt{{{\left( ay \right)}^{2}}+{{\left( bx \right)}^{2}}} \\\ & AC=\sqrt{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}} \\\ \end{aligned}$$ Hence, we can write $$\sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}$$ and $$\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}$$ as $$\begin{aligned} & \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{aligned}$$ Now, we can put values of $$\sin \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}$$ and $$\cos \dfrac{{{\theta }_{1}}+{{\theta }_{2}}}{2}$$ (calculated in equation (9)) in equation (6). $$\begin{aligned} & \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{aligned}$$ Now, we have already given that difference between eccentric angles is 120 i.e. $${{\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{aligned} & \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{aligned}$$ Now, find out the intersection of equation (1) and equation (2), then suppose that intersection as h and k and try to get relationship between h and k by elimination of $${{\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$$ .