10. $$ 9 x^{2}+4 y^{2}-54 x+40 y+37=0 $$

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We start with the equation

$$
9 x^{2}+4 y^{2}-54 x+40 y+37=0.
$$

**Step 1. Group and factor the $$x$$ and $$y$$ terms.**

For $$x$$, factor out 9:

$$
9 x^{2}-54 x = 9\left(x^{2}-6x\right).
$$

For $$y$$, factor out 4:

$$
4 y^{2}+40 y = 4\left(y^{2}+10y\right).
$$

Thus, the equation becomes

$$
9\left(x^{2}-6x\right) + 4\left(y^{2}+10y\right) + 37 = 0.
$$

**Step 2. Complete the square for the $$x$$-terms.**

For $$x^{2}-6x$$, take half of $$-6$$ (which is $$-3$$), and add and subtract its square ($$9$$):

$$
x^{2}-6x = (x-3)^{2} - 9.
$$

Thus,

$$
9\left(x^{2}-6x\right) = 9\left[(x-3)^{2} - 9\right] = 9(x-3)^{2} - 81.
$$

**Step 3. Complete the square for the $$y$$-terms.**

For $$y^{2}+10y$$, take half of $$10$$ (which is $$5$$), and add and subtract its square ($$25$$):

$$
y^{2}+10y = (y+5)^{2} - 25.
$$

Thus,

$$
4\left(y^{2}+10y\right) = 4\left[(y+5)^{2} - 25\right] = 4(y+5)^{2} - 100.
$$

**Step 4. Substitute back into the equation and simplify.**

Replace the completed squares into the equation:

$$
9(x-3)^{2} - 81 + 4(y+5)^{2} - 100 + 37 = 0.
$$

Combine the constant terms:

$$
-81 - 100 + 37 = -144.
$$

Thus, the equation becomes

$$
9(x-3)^{2} + 4(y+5)^{2} - 144 = 0.
$$

Add $$144$$ to both sides:

$$
9(x-3)^{2} + 4(y+5)^{2} = 144.
$$

**Step 5. Write the equation in standard form of an ellipse.**

Divide both sides by $$144$$:

$$
\frac{9(x-3)^{2}}{144} + \frac{4(y+5)^{2}}{144} = 1.
$$

Simplify the fractions:

$$
\frac{(x-3)^{2}}{16} + \frac{(y+5)^{2}}{36} = 1.
$$

This is the standard form of an ellipse with:

- Center at $$(3,-5)$$,
- Semi-minor axis $$a = 4$$ (since $$16 = 4^2$$),
- Semi-major axis $$b = 6$$ (since $$36 = 6^2$$).
The equation simplifies to the standard form of an ellipse: $$ \frac{(x-3)^{2}}{16} + \frac{(y+5)^{2}}{36} = 1 $$ This represents an ellipse centered at $$(3, -5)$$ with a semi-minor axis of 4 and a semi-major axis of 6.

10. \( 9 x^{2}+4 y^{2}-54 x+40 y+37=0 \)

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