SAT1000 - P839

Geometry Level pending

As shown above, point FF is the focus of the parabola C:y2=4xC: y^2=4x, and line ll passing through K(1,0)K(-1,0) intersects with CC at point A,BA,B, and A,DA,D are symmetry about the x-axis.

If FAFB=89\overrightarrow{FA} \cdot \overrightarrow{FB}=\dfrac{8}{9}, then find the equation of the incircle of BDK\triangle BDK.

The equation can be expressed as: (xa)2+(yb)2=r2(x-a)^2+(y-b)^2=r^2. Submit 1000(a+b+r)\lfloor 1000(a+b+r) \rfloor.

Have a look at my problem set: SAT 1000 problems


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