A) \[{{t}_{1}}{{t}_{2}}=-1\]
B) \[{{t}_{1}}{{t}_{2}}=1\]
C) \[{{t}_{1}}+{{t}_{2}}=-1\]
D) \[{{t}_{1}}-{{t}_{2}}=1\]
Correct Answer: A
Solution :
\[\frac{(y-2a{{t}_{2}})}{(2a{{t}_{2}}-2a{{t}_{1}})}=\frac{x-at_{2}^{2}}{(at_{2}^{2}-at_{1}^{2})}\]; As focus i.e., (a, 0) lies on it, Þ \[\frac{-2a{{t}_{2}}}{2a({{t}_{2}}-{{t}_{1}})}=\frac{a(1-t_{2}^{2})}{a({{t}_{2}}-{{t}_{1}})({{t}_{2}}+{{t}_{1}})}\] Þ \[-{{t}_{2}}=\frac{(1-t_{2}^{2})}{({{t}_{2}}+{{t}_{1}})}\] Þ \[-t_{2}^{2}-{{t}_{1}}{{t}_{2}}=1-t_{2}^{2}\] Þ \[{{t}_{1}}{{t}_{2}}=-1\].You need to login to perform this action.
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