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$$f(x,y)={10x\over(x^2+4y^2+9)}$$

What is the smallest possible distance between points $(x_0,y_0)$ and $(x_1, y_1)$ such that $f(x_0, y_0) = 0$ and $f(x_1, y_1) = 1$?

$f(x,y)=0 $ $f(x,y)= 1$ gives $(5,\pm2), (1,0),(9,0)$ and $f(x,y)=0$ gives $x=0$ and $x^2+4y^2+9$ is not equal to zero.

but that's as far as I can go; what is the next step in calculating?

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    $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. $\endgroup$ – Lee David Chung Lin Feb 8 '20 at 5:52
  • $\begingroup$ What is the meaning of "[2]" in your question ? $\endgroup$ – Jean Marie Feb 8 '20 at 6:34
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The first constraint is equivalent to $x_0=0$, i.e., $(x_0,y_0)$ belongs to the $y$ axis ; let us call $(L)$ this axis.

The second one to $$x_1^2+4y_1^2+9=10x_1 \ \iff \ (x_1-5)^2+4y_1^2=4^2 \tag{1}$$

which means that $(x_1,y_1)$ belong to an ellipse $(E)$. This ellipse $(E)$ has the $x$ axis as its symmetry axis.

Let us compute its intersections with the $x$ axis : plugging $y_1=0$ in (1) gives

$$(x_1-5)^2=4^2 \ \ \iff \ \ x_1=1 \ \ \text{or} \ \ x_1=9\tag{2} $$

Therefore $(E)$ is entirely to the right of $(L)$.

As a consequence of (2), the closest point of $(E)$ to $(L)$ is

$$(x_1,y_1)=(1,0)$$

and the closest distance is

$$d=1$$

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