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