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Let $m$ be the slope of the normal $L$ drawn at $(1,2)$ to the curve $x=t^2-7 t+7, y=t^2-4 t-10$ and $a x+b y+c=0$ be the equation of the normal L. If G.C.D of $(a, b, c)$ is 1 , then $m(a+b+c)=$
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Verified Answer
The correct answer is:
$5$
$x=\mathrm{t}^2-7 t+7$
$$
\begin{aligned}
& \frac{d x}{d t}=2 t-7 \\
& y=t^2-4 t-10 \\
& \frac{d y}{d t}=2 t-4 \\
& \frac{d y}{d x}=\frac{2 t-4}{2 t-7}
\end{aligned}
$$
$\therefore$ Slope of normal $=\frac{7-2 t}{2 t-4}$
$\because$ Now, at $(1,2)$ :
$$
\therefore x=t^2-7 t+7=1
$$
$$
\begin{aligned}
& t^2-7 t+6=0 \\
& t=1,6 \\
& y=t^2-4 t-10=2 \\
& t^2-4 t-12=0 \\
& t=6,-2
\end{aligned}
$$
$\therefore t=6$ is common solution
$$
\therefore \text { Slope of normal }=\frac{7-2(6)}{2(6)-4}=\frac{-5}{8}
$$
Equation of normal : $y-2=\frac{-5}{8}(x-1)$
$$
\begin{aligned}
& 8 y-16=-5 x+5 \\
& 5 x+8 y-21=0 \\
& a=5, b=8, c=-21 \\
& a+b+c=5+8-21=-8 \\
& \therefore m(a+b+c)=\frac{-5}{8} x-8=5
\end{aligned}
$$
$$
\begin{aligned}
& \frac{d x}{d t}=2 t-7 \\
& y=t^2-4 t-10 \\
& \frac{d y}{d t}=2 t-4 \\
& \frac{d y}{d x}=\frac{2 t-4}{2 t-7}
\end{aligned}
$$
$\therefore$ Slope of normal $=\frac{7-2 t}{2 t-4}$
$\because$ Now, at $(1,2)$ :
$$
\therefore x=t^2-7 t+7=1
$$
$$
\begin{aligned}
& t^2-7 t+6=0 \\
& t=1,6 \\
& y=t^2-4 t-10=2 \\
& t^2-4 t-12=0 \\
& t=6,-2
\end{aligned}
$$
$\therefore t=6$ is common solution
$$
\therefore \text { Slope of normal }=\frac{7-2(6)}{2(6)-4}=\frac{-5}{8}
$$
Equation of normal : $y-2=\frac{-5}{8}(x-1)$
$$
\begin{aligned}
& 8 y-16=-5 x+5 \\
& 5 x+8 y-21=0 \\
& a=5, b=8, c=-21 \\
& a+b+c=5+8-21=-8 \\
& \therefore m(a+b+c)=\frac{-5}{8} x-8=5
\end{aligned}
$$
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