question_answer

Two mutually perpendicular tangents of the parabola \[{{y}^{2}}=4ax\]meet its axis in \[{{P}_{1}}\]and \[{{P}_{2}}\]If S is the focus of the parabola, then \[\frac{1}{S{{P}_{1}}}+\frac{1}{S{{P}_{2}}}\] is equal to

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, \[S{{P}_{2}}=a+at_{2}^{2}\]

\[\Rightarrow \]\[S{{P}_{2}}=a+a{{\left( \frac{1}{{{t}_{1}}} \right)}^{2}}\]\[\left[ \because {{t}_{2}}=-\frac{1}{{{t}_{1}}} \right]\]\[\Rightarrow \]\[\frac{1}{S{{P}_{1}}}+\frac{1}{S{{P}_{2}}}=\frac{1}{a+at_{1}^{2}}+\frac{t_{1}^{2}}{a{{t}_{1}}7a}\]\[=\frac{1}{a}\left( \frac{1+t_{1}^{2}}{1+t_{1}^{2}} \right)=\frac{1}{a}\]