What relationship exists between the numbers $a$ and $b$, if the ellipse given by the equation $(x / a)^{2}+(y / b)^{2}=1$ is tangent to the hyperbola given by the equation $y=1 / x$?
I. solution. We say that two curves touch each other if they have a common point and in this point they have a common tangent, that is, their (variable) tangent at the common point is equal.

For the $x_{k}$ abscissas of the common points of our curves, eliminating $y$ from the two equations yields:

$$
\begin{gathered}
\frac{x_{k}^{2}}{a^{2}}+\frac{1}{b^{2} x_{k}^{2}}=1 \\
x_{k}^{4}-a^{2} x_{k}^{2}+\frac{a^{2}}{b^{2}}=0
\end{gathered}
$$

(multiplying by $x_{k}^{2}$ is allowed, since $x_{k} \neq 0$, because our hyperbola has no point with $x=0$ abscissa). From here,

$$
x_{k}^{2}=\frac{a^{2}}{2} \pm g, \quad x_{k}= \pm \sqrt{\frac{a^{2}}{2} \pm g}
$$

where

$$
g=\sqrt{\frac{a^{4}}{4}-\frac{a^{2}}{b^{2}}}
$$

Both of our curves are centrally symmetric about the origin, so their points of tangency are also mirror images of each other in pairs, which is why it is enough to ensure the equality of the tangents' slopes at the common points with positive abscissas. For $x_{k}>0$, the hyperbola gives $y>0$, so we express the ordinate of the point on the upper arc of the ellipse from the ellipse equation for the purpose of differentiation:

$$
y=\frac{b}{a} \sqrt{a^{2}-x^{2}}, \quad y^{\prime}=-\frac{b}{a} \cdot \frac{x}{\sqrt{a^{2}-x^{2}}}
$$

Accordingly, at the common point, $x=x_{k}$ and substituting the obtained expressions, the slope of the tangent [after rationalizing the denominator and taking (1) into account] is

$$
-\frac{b}{a} \cdot \frac{\sqrt{\frac{a^{2}}{2} \pm g}}{\sqrt{\frac{a^{2}}{2} \mp g}}=-\frac{b}{a} \cdot \frac{\sqrt{\left(\frac{a^{2}}{2}\right)^{2}-g^{2}}}{\frac{a^{2}}{2} \mp g}=-\frac{1}{\frac{a^{2} \mp g}{2}}
$$

The slope of the tangent to the hyperbola is similarly

$$
y=\frac{1}{x}, \quad y^{\prime}=-\frac{1}{x^{2}}, \quad-\frac{1}{x_{k}^{2}}=-\frac{1}{\frac{a^{2}}{2} \pm g}
$$

and equating this with the previous one, we see that the tangent can only be common at their common point if

$$
g=0, \quad \text { that is } \quad \frac{a^{4}}{4}-\frac{a^{2}}{b^{2}}=0, \quad a b=2
$$

(since $a$ and $b$ are naturally positive).

![](https://cdn.mathpix.com/cropped/2024_05_02_5efae15b136aaf7436afg-1.jpg?height=911&width=895&top_left_y=1797&top_left_x=604)

This gives us the desired relationship between $a$ and $b$. If this condition is satisfied, our curves have 2 common points: $(a / \sqrt{2}, b / \sqrt{2})$ and $(-a / \sqrt{2},-b / \sqrt{2})$, but the ellipse is not uniquely determined, it is a series of ellipses depending on the parameter $a$ with the equation

$$
\frac{x^{2}}{a^{2}}+\frac{a^{2} y^{2}}{4}=1
$$

each of which touches the fixed $y=1 / x$ hyperbola at the point of tangency given by the parameter.

II. solution. It is known from the textbook that the tangent to the $y=1 / x$ hyperbola at the point $\left(x_{0} ; 1 / x_{0}\right)$ intersects the $x$ axis at the point $\left(2 x_{0} ; 0\right)$. Furthermore, that our ellipse is the image of the circle $x^{2}+y^{2}=a^{2}$ in a perpendicular affinity whose axis is the $x$ axis $[1$

It can be proven that the tangents at a corresponding point pair of the circle and the ellipse are also affine images of each other, and thus intersect each other on the axis of the affinity. According to this, the $x_{k}$ abscissa of the common point of the ellipse and the hyperbola is characterized by the fact that the tangent to the circle $x^{2}+y^{2}=a^{2}$ at the point with abscissa $x_{k}$ intersects the $x$ axis at the point $\left(2 x_{k} ; 0\right)$. Now, the equation of the tangent is

$$
x_{k} \cdot x+\sqrt{a^{2}-x_{k}^{2}} \cdot y=a^{2},
$$

and since the coordinates of the said point satisfy this, $2 x_{k}^{2}=a^{2}, x_{k}^{2}=a^{2} / 2$. Thus, the ordinate of the common point, from the hyperbola,

$$
y_{k}=\frac{1}{x_{k}}=\frac{\sqrt{2}}{a}
$$

and from the ellipse,

$$
y_{k}=\frac{b}{a} \cdot \sqrt{a^{2}-\frac{a^{2}}{2}}=\frac{b}{\sqrt{2}}
$$

and from the equality of these two, $a b=2$.

Remark. The tangency of curves is not defined in the textbook. Most solvers arrived at the goal by eliminating one coordinate from the equations of the ellipse and hyperbola, and writing the condition that the equation obtained for the other coordinate has a multiple root, its discriminant is 0, thus at least two common points coincide. The previous textbook examined the condition for the tangency of a parabola and a line in this way.

We accepted such solutions, but we did not share them, because this principle cannot be used if not both curves have an algebraic equation and also not if they are of higher degree. The tools used here, however, are known from the curriculum, the textbook defines the tangency of a curve and a line (a first-degree curve) precisely in the above way.

[^0]:    ${ }^{1}$ Czapáry E.-Horvay K.-Pálmay L.: Mathematics for secondary and vocational schools, 3rd grade. Tankönyvkiadó, Budapest. 1968. p. 318.

    ${ }^{2}$ There, p. 168.

    ${ }^{3}$ There, on p. 198, in the solution of problem 133.

    ${ }^{4}$ There, p. 294.