Equation of ellipse tangent to axes 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.

 A: 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 $h^2 < \frac1{144}$. 
A: In homogeneous coordinates, for the origin and the line $\mathscr l=(4,0,1)\times(0,3,1)=(-3,-4,12)$ to be a polar pair, the last row and column of the matrix $Q$ of the conic must be a scalar multiple of $\mathscr l$, i.e., $$Q=\pmatrix{a&-b&-3\\-b&c&-4\\-3&-4&12}.$$ We also have the tangency constraints $$(4,0,1)Q=(4a-3,-4b-4,0)=\lambda(0,1,0) \\ (0,3,1)Q=(-3b-3,3c-4,0)=\mu(1,0,0)$$ from which $a=\frac34$, $c=\frac43$ and $b\ne-1$, so an equation for this family of conics is $$\pmatrix{x&y&1}\pmatrix{\frac34&-b&-3\\-b&\frac43&-4\\-3&-4&12}\pmatrix{x\\y\\1} = \frac34x^2-2bxy+\frac43y^2-6x-8y+12=0.$$ The discriminant is equal to $4b^2-4$, so this is an ellipse for $b^2\lt1$. FWIW, the centers of these ellipses lie on the ray $\left(\frac4{b+1},\frac3{b+1}\right)$. At $b=-1$, the curve degenerates into a line through the two points, while $b=1$ gives a parabola.  
More generally, if the ellipse is tangent to the coordinate axes at $(h,0)$ and $(0,k)$, ($h,k\ne0$) we will have $$Q=\pmatrix{\frac kh&-b&-k\\-b&\frac hk&-h\\-k&-h&hk}$$ with $b^2\lt1$. Dividing through by $hk\ne0$ produces the perhaps more interesting form $$\pmatrix{\frac1{h^2}&-\frac b{hk}&-\frac1h\\-\frac b{hk}&\frac1{k^2}&-\frac1k\\-\frac1h&-\frac1k&1}.$$
