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If the normal to the parabola $y^2=4 x$ at $P(1,2)$ meets the parabola again in $Q$, then coordinates of $Q$ are
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The correct answer is:
$(9,-6)$
The given equation of the parabola is
$y^2=4 x$
Equation of tangent at $(1,2)$ is
$\begin{aligned} \Rightarrow & & y \cdot 2 & =2(x+1) \\ \Rightarrow & & y & =x+1\end{aligned}$
Slope of tangent $=1$
$\therefore$ slope of the normal $=-1$
Equation of the normal passing through $(1,2)$ is
$\begin{aligned} & y-2=-1(x-1) \\ & y-2=-x+1 \Rightarrow x+y=3\end{aligned}$
This equation again meet at $Q$ of the parabola, then
$\begin{array}{cc} & y^2=4(3-y) \\ \Rightarrow & y^2=12-4 y \Rightarrow y^2+4 y-12=0 \\ \Rightarrow & (y-2)(y+6)=0 \\ \Rightarrow & y=2,6\end{array}$
If $y=-6$, then $x=3-(-6)=9$
$\therefore$ The coordinates of point $Q$ are $(9,-6)$.
$y^2=4 x$
Equation of tangent at $(1,2)$ is
$\begin{aligned} \Rightarrow & & y \cdot 2 & =2(x+1) \\ \Rightarrow & & y & =x+1\end{aligned}$
Slope of tangent $=1$
$\therefore$ slope of the normal $=-1$
Equation of the normal passing through $(1,2)$ is
$\begin{aligned} & y-2=-1(x-1) \\ & y-2=-x+1 \Rightarrow x+y=3\end{aligned}$
This equation again meet at $Q$ of the parabola, then
$\begin{array}{cc} & y^2=4(3-y) \\ \Rightarrow & y^2=12-4 y \Rightarrow y^2+4 y-12=0 \\ \Rightarrow & (y-2)(y+6)=0 \\ \Rightarrow & y=2,6\end{array}$
If $y=-6$, then $x=3-(-6)=9$
$\therefore$ The coordinates of point $Q$ are $(9,-6)$.
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