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Find equation of sketched ellipse touching the coordinate axes at $(0,3), (4,0).$

Geogebra gives it approximately

enter image description here

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.

enter image description here

Hypergeometricx
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Narasimham
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    Are you sure it's unique? With the five points, yes, but I think there's an ellipse with those points of tangency whose axes are parallel to the coordinate axes. – pjs36 May 31 '17 at 03:36
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    A conic is determined by five points. A point plus tangent direction counts as two points, so you have one degree of freedom in choosing your ellipse. Have you any other information about it? – Angina Seng May 31 '17 at 03:38
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    But there's still something you're not mentioning. From my perspective, you've thrown five points down on Geogebra and are asking for the equation of the ellipse even though Geogebra has given it. Perhaps trying to add some motivation for the problem would help (or instead, ask about the family of ellipses like this, but still motivation would help). – pjs36 May 31 '17 at 04:01
  • Holding nothing back,Posted because unable to arrive at it in a quick guess or attempt, hoping it would likewise raise curiosity of someone here.Geogebra was just as an aid, as a relevant example. – Narasimham May 31 '17 at 04:13
  • So finally it turns out as $ (x/a - 1)^2 + (y/b - 1)^2 + 2 h (x /a) * (y/b) = 1, |h|<1 $ – Narasimham May 31 '17 at 11:40
  • This is an interesting question. Here's a graphical implementation of this problem on desmos. I wonder if there is a more direct method of deriving the equation of an ellipse given two tangent points, similar to using Bezier curves for a parabola. Note that when $h=1$, the ellipse is an upright one centre at $(a,b)$. – Hypergeometricx Jun 02 '17 at 13:14

2 Answers2

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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 $h^2 < \frac1{144}$.

Macavity
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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 $$\pmatrix{\frac1{h^2}&-\frac b{hk}&-\frac1h\\-\frac b{hk}&\frac1{k^2}&-\frac1k\\-\frac1h&-\frac1k&1}.$$

amd
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