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If all the normals drawn to the curve $y=\frac{1+3 x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3 x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha, \beta)$, then $3 \alpha+2 \beta=$
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Verified Answer
The correct answer is:
4
$\because y=\frac{1+3 x^2}{3+x^2}$
To get the intersection point, let us solve
$$
\begin{aligned}
& y=\frac{1+3 x^2}{3+x^2} \& y=1: \\
& \frac{1+3 x^2}{3+x^2}=1 \Rightarrow 3 x^2+1=x^2+3 \Rightarrow x= \pm 1 \\
& \frac{d y}{d x}=\frac{16 x}{\left(3+x^2\right)^2}
\end{aligned}
$$
Slope of the normal : $m=-\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{-\left(3+x^2\right)^2}{16 x}$
$\mathrm{Eq}^{\mathrm{n}}$ of normla is
Case 1: $x=1$
Then, $(y-1)=-1(x-1) \Rightarrow y+x=2$
Case $2: \mathrm{x}=-1$
Then $y-1=1(x+1) \Rightarrow-x+y=2$
Solving eqn (i) \& (ii), we get
$$
\mathrm{x}=0 \quad \& \mathrm{y}=2
$$
Then $(\alpha, \beta)=(0,2)$
$$
\therefore 3 \alpha+2 \beta=3 \times 0+2 \times 2=4
$$
To get the intersection point, let us solve
$$
\begin{aligned}
& y=\frac{1+3 x^2}{3+x^2} \& y=1: \\
& \frac{1+3 x^2}{3+x^2}=1 \Rightarrow 3 x^2+1=x^2+3 \Rightarrow x= \pm 1 \\
& \frac{d y}{d x}=\frac{16 x}{\left(3+x^2\right)^2}
\end{aligned}
$$
Slope of the normal : $m=-\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{-\left(3+x^2\right)^2}{16 x}$
$\mathrm{Eq}^{\mathrm{n}}$ of normla is
Case 1: $x=1$
Then, $(y-1)=-1(x-1) \Rightarrow y+x=2$
Case $2: \mathrm{x}=-1$
Then $y-1=1(x+1) \Rightarrow-x+y=2$
Solving eqn (i) \& (ii), we get
$$
\mathrm{x}=0 \quad \& \mathrm{y}=2
$$
Then $(\alpha, \beta)=(0,2)$
$$
\therefore 3 \alpha+2 \beta=3 \times 0+2 \times 2=4
$$
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