Htykuut

Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$

My question is - why partial differentiating makes work easy and what those two equations represent?

You have some errors in your partial derivatives:

We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$

$(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$

$(2)$ Finding the partial derivative with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$ So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$. In the second partial, w.r.t. y, your $us$ should be $hx.$

That gives us the equations of two lines:

$$begin{align} ax + hy+g &= 0 \ \
hx+by +f &= 0end{align}$$

Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.

Indeed, sollving the system gives us the intersection of the partial derivatives at $$left(dfrac{hf-bg}{ab - h^2}right ) , left ( dfrac {gh-af}{ab - h^2}right)$$ provided $ab-h^2 leq 0$ and $bneq 0$.