A) \[\frac{4}{a}\]
B) \[\frac{2}{a}\]
C) \[\frac{1}{a}\]
D) \[\frac{1}{4a}\]
Correct Answer: C
Solution :
Tangents at\[{{P}_{1}}\] and \[{{P}_{2}}\]are mutually perpendiculars |
\[\therefore \] \[{{t}_{1}}{{t}_{2}}=-\,1\] |
Similarly, \[S{{P}_{1}}=\sqrt{{{(a-at_{1}^{2})}^{2}}+{{(2a{{t}_{1}})}^{2}}}\] |
\[S{{P}_{1}}=a+a_{1}^{2}\] |
Similarly, \[S{{P}_{2}}=a+at_{2}^{2}\] |
\[\Rightarrow \]\[S{{P}_{2}}=a+a{{\left( \frac{1}{{{t}_{1}}} \right)}^{2}}\]\[\left[ \because {{t}_{2}}=-\frac{1}{{{t}_{1}}} \right]\]\[\Rightarrow \]\[\frac{1}{S{{P}_{1}}}+\frac{1}{S{{P}_{2}}}=\frac{1}{a+at_{1}^{2}}+\frac{t_{1}^{2}}{a{{t}_{1}}7a}\]\[=\frac{1}{a}\left( \frac{1+t_{1}^{2}}{1+t_{1}^{2}} \right)=\frac{1}{a}\] |
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