We have our general equation is second degree in two variables: $ax^2+2hxy+by^2+2gx+2fy+c=0$
Let's say this represents a pair of straight lines
Our professor told that we could find the point of intersection by partially differentiating it twice, once with x and once with y and solve those 2 equations.
$2ax+2hy+2g=0$
$2hx+2by+2f=0$
I don't really understand how partial derivative of the equation has any significance here. Why is this true? What do the partially differentiated equations represent?
Suppose the two lines given to us are $L(x,y)=0$ and $J(x,y)=0$, where $L(x,y)=px+qy+r$ and $J(x,y)=sx+ty+z$. Then the equation of the pairs of these lines is $$LJ=0. \tag{1}$$ Let us consider the line given by $$L+\lambda J=0. \tag{2}$$ For different values of $\lambda$, we will get a line that passes through the intersection of the lines $L$ and $J$. The only point $(x,y)$ for which (2) will have a solution for every $\lambda$ is when $(x,y)$ is the intersection of the lines $L$ and $J$.
Coming back to equation (1), if we take the partial derivative with regards to $x$ and $y$, then we get \begin{align*} L_x\, J+J_x \, L & =0\\ L_y\, J+J_y \, L & =0. \end{align*} Note that $L_x,L_y,J_x,J_y$ are all constants (real numbers). Suppose say $J_x, J_y \neq 0$, then we can divide by $J_x$ and $J_y$ to get \begin{align*} L +\lambda_1 J & =0\\ L +\lambda_2 J & =0. \end{align*} Which is basically in the form of equation (2). So a common solution for this system will give us the solution for equation (2) that can work for all values of $\lambda$, and hence the intersection point.