221. Find the coordinates of the vertex of the parabola given by the equation $x=2 y^{2}+4 y-5$. Write the equation of the axis of symmetry and plot the parabola.
Solution. Let's find the coordinates of the vertex of the parabola:

$$
x=2\left(y^{2}+2 y-\frac{5}{2}\right)=2\left[(y+1)^{2}-1-\frac{5}{2}\right]=2(y+1)^{2}-7
$$

from which $x+7=2(y+1)^{2}$ or $X=2 Y^{2}$, where $X=x+7$ and $Y=y+1$, i.e., $\alpha=-7, \beta=-1$ (see formulas (4)). Therefore, the vertex of the parabola lies at the point $O_{1}(-7 ;-1)$. The equation of the axis of the parabola is $y=-1$.

We will construct the parabola $X=2 Y^{2}$ in the new coordinate system $X O_{1} Y$ (Fig. 27). Note that it is useful to find the points of intersection of the parabola with the old coordinate axes when constructing it. Let's find them. Let $x=0$, then $2 y^{2}+4 y-5=0$. Solving this equation, we get:

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-049.jpg?height=373&width=494&top_left_y=355&top_left_x=626)

Fig. 27

$$
\begin{gathered}
y=\frac{-2 \pm \sqrt{4+10}}{2}=-1 \pm \frac{\sqrt{14}}{2} \\
y_{1}=-1+\frac{\sqrt{14}}{2} \approx-1+3.7=2.7 ; \quad y_{2} \approx-4.7
\end{gathered}
$$

i.e., the parabola intersects the $O y$ axis at points $A\left(0 ; y_{1}\right)$ and $B\left(0 ; y_{2}\right)$. If $y=0$, then $x=-5$, so the parabola intersects the $O x$ axis at the point $C(-5 ; 0)$.