Let's determine the equation of the tangents drawn from the point $P\left(\frac{3}{5} ; \frac{14}{5}\right)$ to the ellipse $4 x^{2}+9 y^{2}=36$.

---

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. 

---

Determine the equation of the tangents drawn from the point $P\left(\frac{3}{5} ; \frac{14}{5}\right)$ to the ellipse $4 x^{2}+9 y^{2}=36$.
I. solution: It is known that the equation of the tangent line at the point $\left(x_{1}, y_{1}\right)$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is

$$
\frac{x_{1} x}{a^{3}}+\frac{y_{1} y}{b^{2}}=1
$$

In the example, $a=3, b=2$, and thus the equation of the tangent line is

$$
\frac{x_{1}}{9} x+\frac{y_{1}}{4} y=1
$$

To calculate the unknown coordinates $x_{1}, y_{1}$ of the point of tangency, two equations are provided by using the fact that, on the one hand, the given point $P$ lies on the tangent, and on the other hand, the point of tangency lies on the ellipse. Therefore,

$$
\frac{3}{5 \cdot 9} x_{1}+\frac{14}{5 \cdot 4} y_{1}=1
$$

and

$$
4 x_{1}^{2}+9 y_{1}^{2}=36
$$

From (1), $y_{1}=\frac{30-2 x_{1}}{21}$, substituting this value into (2) yields

$$
25 x_{1}^{2}-15 x_{1}-108=0
$$

from which

$$
x_{1}^{\prime}=\frac{12}{5} \quad \text { and } \quad x_{1}^{\prime \prime}=-\frac{9}{5}
$$

and thus

$$
y_{1}^{\prime}=\frac{6}{5} \quad \text { and } \quad y_{1}^{\prime \prime}=\frac{8}{5}
$$

Substituting the obtained pairs of values into (e) gives

$$
8 x+9 y=30
$$

or

$$
x-2 y=-5
$$

the equations of the two sought tangents.

Grätzer György (Bp. VI., Kölcsey g. IV. o. t.)

II. solution: Let the equation of the sought tangent be

$$
y=m x+b
$$

The coordinates of the common points of the tangent and the ellipse are obtained by substituting the value of $y$ from the above equation into the equation of the ellipse:

$$
4 x^{2}+9\left(m x+b^{2}\right)=36
$$

or, rearranging,

$$
\left(4+9 m^{2}\right) x^{2}+18 m b x+\left(9 b^{2}-36\right)=0
$$

In the case of tangency, the intersection points coincide, so the discriminant of the quadratic equation (1) is 0.

Thus,

$$
(18 m b)^{2}-4\left(4+9 m^{2}\right)\left(9 b^{2}-36\right)=0
$$

or, factoring and simplifying by 144,

$$
9 m^{2}-b^{2}+4=0
$$

The tangent $y=m x+b$ passes through the point $P$, so $\frac{14}{5}=m \frac{3}{5}+b$, or

$$
3 m+5 b=14
$$

From (2) and (3),

$$
9 m^{2}=(14-5 b)^{2}=b^{2}-4
$$

after rearranging and simplifying,

$$
6 b^{2}-35 b+50=0
$$

from which

$$
b_{1}=\frac{10}{3}, \quad b_{2}=\frac{5}{2}
$$

and thus

$$
m_{1}=-\frac{8}{9}, \quad m_{2}=\frac{1}{2}
$$

The equations of the sought tangents are therefore

$$
y=-\frac{8}{9} x+\frac{10}{3} \quad \text { and } \quad y=\frac{1}{2} x+\frac{5}{2} \text {. }
$$

Plichta Jenô (Mezőkövesd, I. László g. IV. o. t.)