solve
The parameter equation of the straight line passing through P is
x= 1+tcosa
Y= 1+tsina, where t is the parameter.
Substituting the parametric equation of a straight line into the elliptic equation, we can get.
( 1+3sin？ a)t？ +(8sina+2cosa)t+ 1=0
|PP 1|*|PP2|
=|t 1|*|t2|
=|t 1t2|
=| 1/( 1+3sin？ a)|
= 1/| 1+3sin？ a|
∫0 & lt； = sin? a & lt= 1
∴ 1<； = 1+3sin？ a & lt=4
∴ 1/4<； = 1/| 1+3sin？ a | & lt= 1
When a = 0, there is a maximum value = 1.
When a = 90, there is a minimum value = 1/4.
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