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Number of normals drawn from $(-2,2)$ to parabola $y^2-2y-2-1=0$ is?

The answer given is 1. I don't understand how a parabola can be represented by this equation. Is there a way of bringing this to a standard form such as $y^2=4ax$ or $x^2=4ay$? The solution given says these two things: the equation $$ (y-1) = m(x+1) - 2am -m^3 $$ and that $a = 1/2$. I know that $ y = mx - 2am -m^3 $ is the standard equation for a normal to a parabola, but how did they get this one? Also, how is $a=1/2$?

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    $\begingroup$ Are you sure you have your equation correct? $\endgroup$ – glowstonetrees May 23 '18 at 14:32
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    $\begingroup$ Do you perhaps mean something like $y^2-2y-2x-1=0$? $\endgroup$ – glowstonetrees May 23 '18 at 14:32
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    $\begingroup$ Cuz right now, there is no $x$ dependence $\endgroup$ – glowstonetrees May 23 '18 at 14:33
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    $\begingroup$ If you plot the graph of $y^2-2y-3=0$ in the $xy$-plane, you just get two horizontal lines which are the roots of this quadratic (namely $x=3$ and $x=-1$), so I would guess that there is indeed a printing mistake $\endgroup$ – glowstonetrees May 23 '18 at 14:56
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    $\begingroup$ Does it mean normals parallel to an axis, because surely there can be an infinite number of normals for a continuous function? $\endgroup$ – Henry Lee May 23 '18 at 15:20
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The parabola is shifted, off the origin, which is why you have $y \to (y-1)$ and $x \to (x+1)$ in your equation. And once you complete the square, you can pretty quickly see all the parts including $a$:

$$\begin{align} y^2-2y-2x-1&=0 \\ y^2-2y\color{red}{+1}-2x-1&=0\color{red}{+1} \\ (y-1)^2&=2x+2 \\ (y-1)^2&=2(x+1) \\ &=4\left(\color{red}{\frac{1}{2}}\right)(x+1) \\ \end{align}$$

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