1. Tardigrade
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  4. The tangent the normal drawn to the curve y=x2-x+4 at the point p(1,4) cuts the x axis at A B respectively. Then the area of the triangle P A B in square units is

Q. The tangent $\&$ the normal drawn to the curve $y=x^{2}-x+4$ at the point $p(1,4)$ cuts the $x$ axis at $A\, \&\, B$ respectively. Then the area of the triangle $P A B$ in square units is

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Solution:

$y=x^{2}-x+4, \frac{d y}{d x}=2 x-1$
$\left(\frac{d y}{d x}\right)(1.4)=m=2-1=1$
The tangent is $y-y_{1}=m\left(x-x_{1}\right)$
$y-4=1(x-1)$
$x-y-1+4=0$
$x-y+3=0$ ...(1)
The normal is $y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)$
$y-4=-1(x-1)$
$x+y-4-1=0$
$x+y-5=0$ ...(2)
The tgt cuts the $x -$ axis at $A$
$\therefore y=0$ from eqn $(1) x =-3$
$\therefore A=(-3.0)$
The normal cuts the $x$ - axis at $B$
$\therefore y=0$
From eqn $(2) x=5\,\, B=(5,0)$
Area of the $\triangle A P B=\frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ -3 & 0 & 1 \\ 5 & 0 & 1\end{vmatrix}=16$ sq, unit

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