Given coordinates are A=(2,3),B=(1,2),C=(4,3) and P=(t,t2)
Area of △ABC
=12∣∣∣21423233∣∣∣=12|1−5+6|=1
Area of △PAB
=12∣∣∣t21tt232t2∣∣∣=12|3t−2t2+1+t2−2t|=12|−t2+t+1|
Given, the area of △ABC is twice that of △PAB, so
1=2×12|−t2+t+1|⇒−t2+t+1=±1⇒t2−t−1=±1⇒t2−t=0,2
Now, t2−t=0
⇒t=0,1
Also, t2−t−2=0
⇒(t−2)(t+1)=0⇒t=−1,2
Hence, t=1,2 as t>0