question_answer

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 \[{{F}_{1}}\] and \[{{F}_{2}}\] be the foci of the ellipse, then \[{{(P{{F}_{1}}-P{{F}_{2}})}^{2}}=\]

A) \[4{{a}^{2}}\left( 1-\frac{{{b}^{2}}}{{{d}^{2}}} \right)\]

B) \[{{a}^{2}}\left( 1-\frac{{{b}^{2}}}{{{d}^{2}}} \right)\]

C) \[4{{a}^{2}}\left( 1-\frac{{{a}^{2}}}{{{d}^{2}}} \right)\]

D) \[{{b}^{2}}\left( 1-\frac{{{a}^{2}}}{{{d}^{2}}} \right)\]

Correct Answer: A

Solution :

[a] Let the point P be \[(acos\theta ,bsin\theta )\]

The equation of tangent at P is

\[\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\]                     ? (1)

If d be the length of perpendicular from the centre\[C(0,\,\,0)\] of the ellipse to the tangent given by (1) then

\[d=\frac{1}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}}\]

\[\Rightarrow \frac{1}{{{d}^{2}}}=\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}\]

\[\Rightarrow \frac{{{b}^{2}}}{{{d}^{2}}}=\frac{{{b}^{2}}}{{{a}^{2}}}{{\cos }^{2}}\theta +1-{{\cos }^{2}}\theta \]

\[\Rightarrow 1-\frac{{{b}^{2}}}{{{d}^{2}}}=\left( 1-\frac{{{b}^{2}}}{{{a}^{2}}} \right){{\cos }^{2}}\theta ={{e}^{2}}{{\cos }^{2}}\theta \]       ? (2)

Now,

\[{{(P{{F}_{1}}-P{{F}_{2}})}^{2}}={{(2aecos\theta )}^{2}}\]

\[=\,\,\,\,4{{a}^{2}}{{e}^{2}}{{\cos }^{2}}\theta =4{{a}^{2}}\left( 1-\frac{{{b}^{2}}}{{{d}^{2}}} \right)\]