The surface
$z = x^2 + y^2 \tag{1}$
is in fact the level surface of the function
$F(x, y, z) = x^2 + y^2 -z \tag{2}$
on which $F(x, y, z)$ takes the value $0$. Since the gradient of a differentiable function is normal to its level surfaces, we can indeed obtain a normal to (1) by computing the gradient of $F(x, y, z)$ as given by (2). We have
$\nabla F(x, y, z) = (2x, 2y, -1) \tag{3}$
at any point $(x, y, z) \in \Bbb R^3$; thus, at the point $(1, -2, 5)$, which is easily seen to lie in the surface
$F(x, y, z) = x^2 + y^2 - z = 0, \tag{4}$
($1^2 + (-2)^2 -5 = 0$) we have
$\nabla F(1, -2, 5) = (2, -4, -1). \tag{5}$
Now, I'm not exactly sure what it means to represent a line in vectorial form, but an expression for the line normal to the surface $z = x^2 + y^2$ at the point $(1, -2, 5)$ may now be written in terms of the normal vector $\vec n = \nabla F(1, -2, 5)$ and the position vector field $\vec r= (x, y, z)$ on $\Bbb R^3$; taking $\vec r_0 = (1, -2, 5)$, we have
$\vec r(t) = \vec r_0 + t\nabla F(1,2,5), \tag{6}$
or
$\vec r(t) = (1, -2, 5) + t(2, -4, -1). \tag{7}$
An expression for the tangent plane may be had in a roughly similar manner; $\vec r = (x, y, z)$ is a point in the tangent plane if and only if the vector $\vec r - \vec r_0$ lies in that plane and is hence perpendicular to $\nabla F(1, -2, 5)$; thus we may write
$(\vec r - \vec r_0) \cdot \nabla F(1, -2, 5) = (\vec r - \vec r_0) \cdot (2, -4, -1) =0 \tag{8}$
for the equation of the (tangent) plane passing through the point $\vec r_0 = (1, -2, 5)$. A little algebra allows (8) to be re-written in another well-known form:
$2x -4y - z = \vec r_0 \cdot (2, -4, -1) = (1, -2, 5) \cdot (2, -4, -1) = 5. \tag{9}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Best Answer
You can find the slope of the tangent line by differentiating. If you don't want to do implicit differentiation (which may be simpler in this case), you can just do some algebra beforehand: $$x^5+y^5=2x^3 \to y^5=2x^3-x^5 \to y=\sqrt[5]{2x^3-x^5}$$ and differentiate according to the power rule. You have the slope of your tangent line; knowing that it goes through $(1,1)$, you should have enough information to solve for that.
The tangent vector will have a slope exactly the same as that of the tangent line. The normal vector will have a slope that is the negative inverse of that of the tangent vector. If $m_t$ is the slope of the tangent vector, the slope $m_n$ of the normal vector will be $-\frac{1}{m_t}$.
In the method you mentioned, you really just have to find $c$ here to match it up with that equation. You would just re-arrange the equation so that all the constants are on one side of the equation. Here, we find that $$F(x,y)=x^5-2x^3+y^5=c$$ and $c=0$.