If P lies in the first quadrant on the ellips | vedclass

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(A) Let $PF_1 = r_1$ and $PF_2 = r_2$. The tangent at $P$ is the external angle bisector of $\angle F_1PF_2$,and the normal at $P$ is the internal ang

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(A) Let $PF_1 = r_1$ and $PF_2 = r_2$. The tangent at $P$ is the external angle bisector of $\angle F_1PF_2$,and the normal at $P$ is the internal angle bisector of $\angle F_1PF_2$.By the Angle Bisector Theorem,the tangent at $P$ divides the line segment $F_1F_2$ externally in the ratio $r_2 : r_1$,so $\frac{\left| F_2T \right|}{\left| F_1T \right|} = \frac{r_2}{r_1}$.The normal at $P$ divides the line segment $F_1F_2$ internally in the ratio $r_2 : r_1$,so $\frac{\left| F_2N \right|}{\left| F_1N \right|} = \frac{r_2}{r_1}$.Thus,$\frac{\left| F_2N \right|}{\left| F_1N \right|} = \frac{\left| F_2T \right|}{\left| F_1T \right|} = \frac{r_2}{r_1}$.Applying componendo and dividendo to both ratios:$\frac{\left| F_2N \right| + \left| F_1N \right|}{\left| F_2N \right| - \left| F_1N \right|} = \frac{r_2 + r_1}{r_2 - r_1}$ and $\frac{\left| F_2T \right| - \left| F_1T \right|}{\left| F_2T \right| + \left| F_1T \right|} = \frac{r_2 - r_1}{r_2 + r_1}$.Multiplying these two expressions:$\left( \frac{\left| F_2N \right| + \left| F_1N \right|}{\left| F_2N \right| - \left| F_1N \right|} \right) \left( \frac{\left| F_2T \right| - \left| F_1T \right|}{\left| F_2T \right| + \left| F_1T \right|} \right) = \left( \frac{r_2 + r_1}{r_2 - r_1} \right) \left( \frac{r_2 - r_1}{r_2 + r_1} \right) = 1$.

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