A) 4
B) 32
C) 8
D) 16
Correct Answer: D
Solution :
Given equation of curve is \[y={{x}^{2}}-x+4\] Slope of tangent at \[P(1,4)\] is \[\left( \frac{dy}{dx} \right)=2x-1\] \[\Rightarrow \] s\[{{\left( \frac{dy}{dx} \right)}_{(1,4)}}=2-1=1\] \[\therefore \] Equation of tangent is \[y-4=1(x-1)\] \[\Rightarrow \] \[y-x=3\] ....(i) and equation of normal at point \[P(1,4)\] is \[y-4=-1(x-1)\] \[\Rightarrow \]\[x+y=5\] ...(ii) Since the tangent cuts x-axis at A \[\therefore \] Co-ordinates of A are \[(-3,0)\] and the normal cuts x-axis at B \[\therefore \] Co-ordinates of Bare \[(5,0)\] Area of \[\Delta PAB\] \[=\frac{1}{2}\left\| \begin{matrix} 1 & 4 & 1 \\ -3 & 0 & 1 \\ 5 & 0 & 1 \\ \end{matrix} \right\|\] \[=\frac{1}{2}|[-4(-3-5)]|=\frac{1}{2}|32|\] \[=16sq\] unitYou need to login to perform this action.
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