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If the tangent drawn to the curve $\left(x^2+1\right)(y-3)=x$ at a point $\mathrm{P}$, lying in the first quadrant, is a horizontal line, then the equation of the normal at the point $\mathrm{P}$ is
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The correct answer is:
$x=1$
$\because\left(x^2+1\right)(y-3)=x$... (i)
$\begin{aligned} & \text { Then, }\left(x^2+1\right) \frac{d y}{d x}+(2 x+0)(y-3)=1 \\ & \Rightarrow \frac{d y}{d x}=\frac{6 x-2 x y+1}{x^2+1}\end{aligned}$
$\because$ Tangent is a horizontal line.
$\because \frac{d y}{d x}=0 \Rightarrow 6 x-2 x y+1=0$ ...(ii)
Solving equations (i) and (ii), we get
$x=1, y=\frac{7}{2}$
Slope of normal, $m=-\frac{1}{\frac{d y}{d x}}=-\frac{1}{0}$
Equation of normal at $\left(1, \frac{7}{2}\right)$ is
$\left(y-\frac{7}{2}\right)=\left(-\frac{1}{0}\right)(x-1) \Rightarrow(x-1)=0$
$\Rightarrow x=1$
$\begin{aligned} & \text { Then, }\left(x^2+1\right) \frac{d y}{d x}+(2 x+0)(y-3)=1 \\ & \Rightarrow \frac{d y}{d x}=\frac{6 x-2 x y+1}{x^2+1}\end{aligned}$
$\because$ Tangent is a horizontal line.
$\because \frac{d y}{d x}=0 \Rightarrow 6 x-2 x y+1=0$ ...(ii)
Solving equations (i) and (ii), we get
$x=1, y=\frac{7}{2}$
Slope of normal, $m=-\frac{1}{\frac{d y}{d x}}=-\frac{1}{0}$
Equation of normal at $\left(1, \frac{7}{2}\right)$ is
$\left(y-\frac{7}{2}\right)=\left(-\frac{1}{0}\right)(x-1) \Rightarrow(x-1)=0$
$\Rightarrow x=1$
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