A $4 x^{2}+9 y^{2}=36$ ellipse which point is closest and which is farthest from the line $2 y=x+7$?
The task will first be solved in a general manner. Let the given equations be:

$$
A x+B y+C=0
$$

and

$$
b^{2} x^{2}+a^{2} y^{2}-a^{2} b^{2}=0
$$

Let the coordinates of the point satisfying our task be $x_{0}$ and $y_{0}$, then the distance of this point from the line (1) is $d$, assuming that $C$ is not positive:

$$
d=\frac{A x_{0}+B y_{0}+C}{\sqrt{A^{2}+B^{2}}}
$$

When does $d$ reach its extreme value? At the same time, when the function

$$
F=d \sqrt{A^{2}+B^{2}}-C=A x_{0}+B y_{0}
$$

The task is to find the value of $x_{0}, y_{0}$ for which the function

$$
F=A x_{0}+B y_{0}
$$

reaches its extreme value, given that:

$$
b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}=0
$$

From (5):

$$
y_{0}= \pm \frac{b}{a} \sqrt{a^{2}-x_{0}^{2}}
$$

thus

$$
F=A x_{0} \pm \frac{b B}{a} \sqrt{a^{2}-x_{0}^{2}}=a x_{0}+\beta \sqrt{\gamma-x^{2}}
$$

from which:

$$
F^{2}=a^{2} x_{0}^{2} \pm 2 \alpha \beta x_{0} \sqrt{\gamma-x_{0}^{2}}-\beta^{2} x^{2}+\beta^{2} \gamma
$$

Add the following equality to this equation:

$$
\left[\beta x_{0} \mp \alpha \sqrt{-x_{0}^{2}+\gamma}\right]^{2}=\beta^{2} x^{2} \mp 2 \alpha \beta x_{0} \sqrt{\gamma-x^{2}}-\alpha^{2} x^{2}+\alpha^{2} \gamma
$$

then:

$$
F^{2}+\left[\beta x_{0} \mp \alpha \sqrt{\gamma-x_{0}^{2}}\right]^{2}=\gamma\left(\alpha^{2}+\beta^{2}\right)
$$

from which

$$
F= \pm \sqrt{\gamma\left(\alpha^{2}+\beta^{2}\right)-\left[\beta x_{0} \mp \alpha \sqrt{\gamma-x_{0}^{2}}\right]^{2}}
$$

$F$ thus reaches its extreme value when

$$
\beta x_{0} \mp \alpha \sqrt{\gamma-x_{0}^{2}}=0
$$

that is, when:

$$
x_{0}=\frac{ \pm \alpha \sqrt{\gamma}}{\sqrt{\alpha^{2}+\beta^{2}}}=\frac{ \pm A a^{2}}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}}
$$

The corresponding $y_{0}$ values are obtained from (5) by substituting the value of $x_{0}$:

$$
y_{0}= \pm \frac{B b^{2}}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}}
$$

So we have two points for which $d$ is at its extreme value, namely

$$
P_{1}\left\{x_{0}^{\prime}=A a^{2} \cdot \frac{1}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}} ; y_{0}^{\prime}=B b^{2} \cdot \frac{1}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}}\right\}
$$

$$
P_{2}\left\{x_{0}^{\prime \prime}=-A a^{2} \cdot \frac{1}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}} ; y_{0}^{\prime \prime}=-B b^{2} \cdot \frac{1}{\sqrt{a^{2} A^{2}+b^{2} B^{2}}}\right\}
$$

In our case

$$
a=3, b=2, A=-1, B=2,
$$

so

$$
P_{1}=\left\{-\frac{9}{5} ; \frac{8}{5}\right\} \quad \text { and } \quad P_{2}=\left\{\frac{9}{5} ;-\frac{8}{5}\right\}
$$

If the distance of the given line from $P_{1}$ is $d_{1}$ and from $P_{2}$ is $d_{2}$, then from (3):

$$
d_{1}=\frac{\left(a^{2} A^{2}+b^{2} B^{2}\right)+C \sqrt{a^{2} A^{2}+b^{2} B^{2}}}{\sqrt{a^{2} A^{2}+b^{2} B^{2}} \cdot \sqrt{A^{2}+B^{2}}}=\sqrt{\frac{a^{2} A^{2}+b^{2} B^{2}}{A^{2}+B^{2}}}+C \frac{1}{\sqrt{A^{2}+B^{2}}}
$$

and

$$
d_{2}=-\sqrt{\frac{a^{2} A^{2}+b^{2} B^{2}}{A^{2}+B^{2}}}+C \frac{1}{\sqrt{A^{2}+B^{2}}}
$$

Now, only positive numbers stand under the square roots, so

$$
\sqrt{\frac{a^{2} A^{2}+b^{2} B^{2}}{A^{2}+B^{2}}}=m>0
$$

and

$$
\frac{1}{\sqrt{A^{2}+B^{2}}}=n>0
$$

$C$ was a negative number according to our condition, so:

$$
-C=c>0
$$

and thus

$$
\begin{gathered}
d_{1}=m-c n \\
d_{2}=-(m+c n)
\end{gathered}
$$

Therefore, the absolute value of $d_{2}$ is greater than the absolute value of $d_{1}$, which means that the point $\left(x_{0}^{\prime}, y_{0}^{\prime}\right)$ corresponds to a minimum, while the point $\left(x_{0}^{\prime \prime}, y_{0}^{\prime \prime}\right)$ corresponds to a maximum.

If we also check the values of $x_{0}$ and $y_{0}$ under $(\alpha)$ and $(\beta)$, we notice that $C$ does not appear in them, meaning the position of points $P_{1}$ and $P_{2}$ is independent of $C$, but depends on $\frac{B}{A}$, because:

$$
x_{0}=\frac{ \pm a^{2}}{\sqrt{a^{2}+b^{2}\left(\frac{B}{A}\right)^{2}}} \quad \text { and } \quad y_{0}=\frac{ \pm b^{2}}{\sqrt{a^{2}+b^{2}\left(\frac{B}{A}\right)^{2}}}
$$

And since $-\frac{B}{A}$ is the slope of the given line, it is clear that every line parallel to the given line corresponds to the same points $P_{1}$ and $P_{2}$. For example, if we draw a line through $P_{1}$ parallel to the given line, which intersects the ellipse at another point $P_{1}^{\prime}$, then the distance of $P_{1}^{\prime}$ from the line (1) would also be $d_{1}$, which would contradict our results; therefore, $P_{1}^{\prime}$ must coincide with $P_{1}$, or in other words, the tangents to the ellipse at points $P_{1}$ and $P_{2}$ are parallel to the given line, which can also be directly proven. For example, the equation of the tangent at $P_{1}$ is:

$$
\frac{x x_{0}^{\prime}}{a^{2}}+\frac{y y_{0}^{\prime}}{b^{2}}=1
$$

from which, substituting the values of $x_{0}^{\prime}, y_{0}^{\prime}$, we get:

$$
A x+B y=\sqrt{a^{2} A^{2}+b^{2} B^{2}},
$$

so the slope of this tangent is also: $-\frac{B}{A}$.

(Antal Márkus.)

Solutions were submitted by: Ádámffy E., Bartók I., Braun I., Dömény E., Dömény I., Enyedi B., Haar A., Kürti I., Liebner A., Neidenbach E., Pichler S., Rosenberg J., Riesz K., Szombathy J., Szúcs A.