Because AE= 1, AD=6.
So ED=5
In RT△DEC
DE=5
EU: DC=4:3
So EC=4
DC=3
In RT△P 1CE.
P 1E=(4+X^2)^ 1/2
ce⊥ef ce⊥cp 1
So CP 1 is parallel to EF.
So s △ EFP1= ef * EC *1/2 = 8.
At △C 1EM and △ EP1n.
Rectangular equation
∠P 1EN=∠EC 1M
EP 1=C 1E
So △C 1EM and △EP 1N are congruent.
And ECP 1N is a rectangle.
So EN=x
So △C 1EF=2x.
△c 1fp 1=△c 1ep 1-△efp 1-△c 1ef= 1/2*x^2-2x-6