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The tangent and the normal drawn to the curve $y=x^{2}-x+4$ at $P(1,4)$ cut the $X$-axis at $A$ and $B$ respectively. If the length of the subtangent drawn to the curve at $P$ is equal to the length of the subnormal, then the area of the $\triangle P A B$ in square units is,
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The correct answer is:
16
Given, equation of curve is
$y=x^{2}-x+4$
Slope of tangent at $P(1,4)$ is
$\frac{d y}{d x}=2 x-1 \Rightarrow\left(\frac{d y}{d x}\right)_{(1,4)}=2-1=1$
So, equation of tangent is
$y-4=1(x-1)$
$\Rightarrow \quad y-x=3...(i)$
and cquation of normal at point $P(1,4)$ is
$\begin{aligned}
&y-4=-1(x-1) \\
&x+y=5 ...(ii)
\end{aligned}$
Since, the tangent cuts $X$-axis at $A$.
Therefore, coordinates of $A$ are $(-3,0)$ and the normal cuts $X$-axis at $B$ and coordinates of $B$ are $(5,0)$.
Therefore, area of $\triangle P A B$
$=\frac{1}{2}\left|\begin{array}{ccc}1 & 4 & 1 \\ -3 & 0 & 1 \\ 5 & 0 & 1\end{array}\right|$
$=\frac{1}{2}[1(0)-4(-3-5)+1(0)]$
$=\frac{1}{2}|32|=16$ sq. units
$y=x^{2}-x+4$
Slope of tangent at $P(1,4)$ is
$\frac{d y}{d x}=2 x-1 \Rightarrow\left(\frac{d y}{d x}\right)_{(1,4)}=2-1=1$
So, equation of tangent is
$y-4=1(x-1)$
$\Rightarrow \quad y-x=3...(i)$
and cquation of normal at point $P(1,4)$ is
$\begin{aligned}
&y-4=-1(x-1) \\
&x+y=5 ...(ii)
\end{aligned}$
Since, the tangent cuts $X$-axis at $A$.
Therefore, coordinates of $A$ are $(-3,0)$ and the normal cuts $X$-axis at $B$ and coordinates of $B$ are $(5,0)$.
Therefore, area of $\triangle P A B$
$=\frac{1}{2}\left|\begin{array}{ccc}1 & 4 & 1 \\ -3 & 0 & 1 \\ 5 & 0 & 1\end{array}\right|$
$=\frac{1}{2}[1(0)-4(-3-5)+1(0)]$
$=\frac{1}{2}|32|=16$ sq. units
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