We first describe the normal computational approach, and then a more conceptual approach.
Computational approach: You can find the equation of the tangent line. After some work you will end up with $y=3-x$.
Substitute $3-x$ for $y$ in the equation of the curve. We end up with a cubic equation $x^3-3x+2=0$.
This equation has $x=1$ as a root. Divide $x^3-3x+2$ by $x-1$. You will get $x^2+x-2$, which factors as $(x-1)(x+2)$. That gives the $x=-2$.
Conceptual approach: Just imagine finding the equation of the tangent line, as we did above, but don't do the actual work. When we substitute for $y$ in the equation of the curve, we end up with a cubic equation of the shape $P(x)=x^3+ax+b=0$.
Since we are not doing the work, we won't find $a$ or $b$.
The tangent line at $x=1$ kisses the curve at $x=1$. Thus the equation $P(x)=0$ has $x=1$ as a double root. So the sum of the roots is $1+1+w$, where $w$ is our mystery number.
The sum of the roots of $P(x)=0$ is the negative of the coefficient of $x^2$, in this case, $0$. So $1+1+w=0$ and therefore $w=-2$.
Given the curve $\mathcal{C}$ as
$$
x^2 + x y - 2 y^2 = 0. \tag 1
$$
Solve the curve $\mathcal{C}$
$$
x^2 + x y - 2 y^2 = 0.\\
\Downarrow\\
4 x^2 + 4 x y - 8 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 - 9 y^2 = 0.\\
\Downarrow\\
\Big( 2 x + y \Big)^2 = \Big( 9 y \Big)^2.\\
\Downarrow\\
2 x + y = \pm 3 y.\\
\Downarrow\\
2 x = \Big( \pm 3 - 1 \Big) y.\\
\Downarrow\\
y = \frac{2}{ \pm 3 - 1 } x.\\
\Downarrow\\
y = \frac{2}{2} x \vee y = \frac{2}{-4} x.\\
$$
So the curve $\mathcal{C}$ are two crossing lines,
given by
$$
y = x \vee y = - \frac{1}{2} x. \tag{2}
$$
The normal through point $(4,4)$
The normal through point $(4,4)$ is given by
$$
y = 8 - x. \tag 3
$$
Intersection
We need to find the intersection between
$$
y = - \frac{1}{2} x
$$
and
$$
y = 8 - x.
$$
So we get
$$
y = - \frac{1}{2} x = 8 - x.\tag 4
$$
Solving intersection
As
$$
- \frac{1}{2} x = 8 - x,
$$
we get
$$
\frac{1}{2} x = 8,
$$
so
$$
x = 16.
$$
And we have
$$
y = - \frac{1}{2} x,
$$
so
$$
y = - 8.
$$
Solution
The solution is given by
$$
\bbox[16px,border:2px solid #800000] { (x,y) = (16,-8). } \tag 5
$$
Best Answer
Let the point of tangency be $(a,a^2)$. Then the slope of the tangent line is $2a$, and the equation of the tangent line is $$y-a^2=2a(x-a).$$ Substitute $x=8$, $y=2$, and solve the resulting quadratic for $a$.