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 \[\left( \frac{dy}{dx} \right)=2x-1\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(1,4)}}=2-1=1\] \[\therefore \]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 \[\therefore \]Coordinates of A are\[(-3,0)\] and the normal cuts\[x-\]axis at B \[\therefore \]Coordinates of Bare (5, 0) \[\therefore \]Area of \[\Delta PAB\] \[=\frac{1}{2}\left| \left| \begin{matrix} 1 & 4 & 1 \\ -3 & 0 & 1 \\ 5 & 0 & 1 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}|[-4(-3-5)]|=\frac{1}{2}|32|\] \[=16\]sq unitYou need to login to perform this action.
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