Example - 1

Let Fand F2 be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that \[\left( {\frac{{\tan \angle P{F_1}{F_2}}}{2}} \right)\tan \left( {\frac{{\angle P{F_2}{F_1}}}{2}} \right) = \frac{{1 - e}}{{1 + e}}\]

Solution: Assume the ellipse to be \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\) and let P be the point \((a\cos \theta ,\,\,b\sin \theta ).\)

Using the sine rule in \(\Delta P{F_1}{F_2},\) we have

\[\begin{align}&\qquad \frac{{P{F_2}}}{{\sin {\phi _1}}} = \frac{{P{F_1}}}{{\sin {\phi _2}}} = \frac{{{F_1}{F_2}}}{{\sin (\pi - ({\phi _1} + {\phi _2}))}} \hfill \\ &\Rightarrow\quad \frac{{P{F_2} + P{F_1}}}{{\sin {\phi _1} + \sin {\phi _2}}} = \frac{{{F_1}{F_2}}}{{\sin ({\phi _1} + {\phi _2})}} \hfill \\ & \Rightarrow\quad\frac{{2a}}{{\sin {\phi _1} + \sin {\phi _2}}} = \frac{{2ae}}{{\sin ({\phi _1} + {\phi _2})}} \hfill \\&\left\{ \begin{gathered}\because \,\,\,P{F_1} + P{F_2}\,\,{\text{will always equal }} \hfill \\{\text{the length of the major axis}} \hfill \\\end{gathered} \right\} \hfill \\&\Rightarrow \quad \frac{{\cos \left( {\frac{{{\phi _1} + {\phi _2}}}{2}} \right)}}{{\cos \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right)}} = e \hfill \\&\Rightarrow \quad \frac{{\cos \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right) - \cos \left( {\frac{{{\phi _1} + {\phi _2}}}{2}} \right)}}{{\cos \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right) + \cos \left( {\frac{{{\phi _1} + {\phi _2}}}{2}} \right)}} = \frac{{1 - e}}{{1 + e}} \hfill \\&\Rightarrow\quad \tan \left( {\frac{{{\phi _1}}}{2}} \right)\tan \left( {\frac{{{\phi _2}}}{2}} \right) = \frac{{1 - e}}{{1 + e}} \hfill \\\end{align} \]

This is the desired result.

Example – 2

Let d be the perpendicular distance from the centre of the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) to the tangent drawn at a point P on the ellipse. If Fand F2 are the two foci of the ellipse, prove that

\[{(P{F_1} - P{F_2})^2} = 4{a^2}\left( {1 - \frac{{{b^2}}}{{{d^2}}}} \right)\]

Solution: Let P be the point \((a\cos \theta ,\,\,b\sin \theta )\) whereas Fand F2 are given by \(( \pm ae,\,0).\)

By definition, the focal distance of any point on an ellipse is e times the distance of that point from the corresponding directrix. Thus,

\[\begin{align}&P{F_1} = \left| {e\left( {a\cos \theta - \frac{a}{e}} \right)} \right|\\&\quad\quad{\rm{ }} = \left| {ae\cos \theta - a} \right|\\&\qquad{\rm{ }} = a - ae\cos \theta \\&\;P{F_2} = e\left( {a\cos \theta + \frac{a}{e}} \right)\\&\quad\quad{\rm{ }} = a + ae\cos \theta \\& \Rightarrow  {(P{F_1} - P{F_2})^2} = 4{a^2}{e^2}{\cos ^2}\theta\qquad\qquad ...\left( 1 \right)\end{align}\]

Now, the equation of the tangent at P is

\[bx\cos \theta + ay\sin \theta - ab = 0\]

The distance of \((0,0)\) from this tangent is d. Thus,

\[\begin{align}&\qquad \qquad d = \frac{{ab}}{{\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}\\& \Rightarrow  \qquad\frac{{{b^2}}}{{{d^2}}} = {\sin ^2}\theta + \frac{{{b^2}}}{{{a^2}}}{\cos ^2}\theta \\& \Rightarrow\qquad  1 - \frac{{{b^2}}}{{{d^2}}} = \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right){\cos ^2}\theta \\&\qquad\qquad\quad\qquad{\rm{ }} = {e^2}{\cos ^2}\theta \\ &\Rightarrow \quad 4{a^2}\left( {1 - \frac{{{b^2}}}{{{d^2}}}} \right) = 4{a^2}{e^2}{\cos ^2}\theta \quad\qquad\qquad ...\left( 2 \right)\end{align}\]

From (1) and (2), we see that the equality stated in the question does indeed hold.

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