# Ellipses Finding the smallest distance

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

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?

• Welcome to MSE. Please edit and use MathJax to properly format math expressions. – Lee David Chung Lin Feb 8 '20 at 5:52
• What is the meaning of "" in your question ? – Jean Marie Feb 8 '20 at 6:34

## 1 Answer 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

$$d=1$$