# Number of normals from (-2,2) for parabola $y^2-2y-2-1=0$?

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 $a=1/2$?

• Are you sure you have your equation correct? – glowstonetrees May 23 '18 at 14:32
• Do you perhaps mean something like $y^2-2y-2x-1=0$? – glowstonetrees May 23 '18 at 14:32
• Cuz right now, there is no $x$ dependence – glowstonetrees May 23 '18 at 14:33
• 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 – glowstonetrees May 23 '18 at 14:56
• Does it mean normals parallel to an axis, because surely there can be an infinite number of normals for a continuous function? – Henry Lee May 23 '18 at 15:20

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}