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If a normal to the parabola $y^2=12 x$ at $A(3,-6)$ cuts the parabola again at $P$, then the equation of the tangent at $P$ is
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The correct answer is:
$x+y=45$
We have,
$y^2=12 x$
$\therefore \quad 2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{6}{y}$
$\therefore$ Slope of Normal at $A(3,-6)=\frac{-1}{\left.\frac{d y}{d x}\right|_{(3,-6)}}=\frac{-1}{\left(\frac{6}{-6}\right)}=1$
$\therefore$ Equation of normal at $A(3,-6)$ is
$y+6=1(x-3) \Rightarrow y=x-9$
On solving $y^2=12 x$ and $y=x-9$, we get $(3,-6)$ and $(27,18)$
$\therefore$ Coordinate of $P$ are $(27,18)$
$\therefore$ Equation of tangent at $P(27,18)$ is
$\begin{aligned} y-18 & =-1(x-27) \\ y-18 & =-x+27 \\ x+y & =45\end{aligned}$
$y^2=12 x$
$\therefore \quad 2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{6}{y}$
$\therefore$ Slope of Normal at $A(3,-6)=\frac{-1}{\left.\frac{d y}{d x}\right|_{(3,-6)}}=\frac{-1}{\left(\frac{6}{-6}\right)}=1$
$\therefore$ Equation of normal at $A(3,-6)$ is
$y+6=1(x-3) \Rightarrow y=x-9$
On solving $y^2=12 x$ and $y=x-9$, we get $(3,-6)$ and $(27,18)$
$\therefore$ Coordinate of $P$ are $(27,18)$
$\therefore$ Equation of tangent at $P(27,18)$ is
$\begin{aligned} y-18 & =-1(x-27) \\ y-18 & =-x+27 \\ x+y & =45\end{aligned}$
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