Find equation of sketched ellipse touching the coordinate axes at $(0,3), (4,0).$
Geogebra gives it approximately
Just as we have for a parabola
$$ \sqrt\frac{x}{a} + \sqrt\frac{y}{b} =1, $$
am trying to find such a corresponding ellipse equation in a symbolic form where $a=4,, b=3 $ is sketched in tangency situation.


It is not hard to take the obvious approach ... start from the generic conic $$ax^2+by^2+2hxy + fx+gy+1=0$$
For an ellipse we need $h^2 < ab$
Setting $x=0$, we have $by^2+gy+1=0$ having a double root at $y=3$, so this is equivalent to $by^2+gy+1 = \frac19(x^2-6x+9)$, whence $b = \frac19, g = -\frac23$.
Similarly, setting $x=0$ we get $ax^2+fx+1=\frac1{16}(x^2-8x+16) \implies a = \frac1{16}, f = -\frac12$.
Simplifying, the ellipse family is $$9x^2 + 16y^2 + 288hxy -72x - 96y+144 = 0$$ parametrised by $h$ where $h^2 < \frac1{144}$.