First, let any point p 1 be C 1M⊥EF, and extend EF to P 1N⊥EF.

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