Discuss the conic section $$ 9 x^{2}-4 y^{2}-36 x-24 y-35=0 $$

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We start with the equation
$$
9x^2 - 4y^2 - 36x - 24y - 35 = 0.
$$

**Step 1. Group Like Terms**

Group the $$x$$ and $$y$$ terms:
$$
9x^2 - 36x - 4y^2 - 24y = 35.
$$
Factor the coefficients of the quadratic terms:
$$
9(x^2 - 4x) - 4(y^2 + 6y) = 35.
$$

**Step 2. Complete the Square**

For the $$x$$-terms, complete the square:
$$
x^2 - 4x = (x-2)^2 - 4.
$$
For the $$y$$-terms, complete the square:
$$
y^2 + 6y = (y+3)^2 - 9.
$$

Substitute back into the equation:
$$
9\Big[(x-2)^2 - 4\Big]- 4\Big[(y+3)^2 - 9\Big] = 35.
$$
Expand:
$$
9(x-2)^2 - 36 - 4(y+3)^2 + 36 = 35.
$$
Notice that $$-36 + 36 = 0$$, so we have:
$$
9(x-2)^2 - 4(y+3)^2 = 35.
$$

**Step 3. Write in Standard Form**

Divide both sides by $$35$$ to obtain:
$$
\frac{9(x-2)^2}{35} - \frac{4(y+3)^2}{35} = 1.
$$
This can be rewritten as:
$$
\frac{(x-2)^2}{\frac{35}{9}} - \frac{(y+3)^2}{\frac{35}{4}} = 1.
$$

**Step 4. Identify the Conic Section**

The standard form
$$
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
$$
represents a hyperbola with center $$(h,k)$$ and a horizontal transverse axis. Comparing, we have:
- Center: $$(2, -3)$$
- $$a^2 = \frac{35}{9}$$ so that $$a = \frac{\sqrt{35}}{3}$$
- $$b^2 = \frac{35}{4}$$ so that $$b = \frac{\sqrt{35}}{2}$$

**Step 5. Additional Features**

- **Vertices:** Located along the horizontal axis relative to the center:
  $$
  (2 \pm a,\,-3) = \left(2 \pm \frac{\sqrt{35}}{3},\, -3\right).
  $$
  
- **Asymptotes:** The equations of the asymptotes for a hyperbola in this form are given by:
  $$
  y+3 = \pm \frac{b}{a}(x-2).
  $$
  Here,
  $$
  \frac{b}{a} = \frac{\frac{\sqrt{35}}{2}}{\frac{\sqrt{35}}{3}} = \frac{3}{2},
  $$
  so the asymptotes are:
  $$
  y+3 = \pm \frac{3}{2}(x-2).
  $$

- **Foci:** The distance from the center to each focus is given by:
  $$
  c = \sqrt{a^2+b^2} = \sqrt{\frac{35}{9}+\frac{35}{4}} = \sqrt{\frac{140+315}{36}} = \sqrt{\frac{455}{36}} = \frac{\sqrt{455}}{6}.
  $$
  Thus, the foci are:
  $$
  \left(2 \pm \frac{\sqrt{455}}{6},\, -3\right).
  $$

**Conclusion**

The given equation represents a hyperbola with center $$(2,-3)$$, horizontal transverse axis, vertices at $$\displaystyle \left(2 \pm \frac{\sqrt{35}}{3}, -3\right)$$, asymptotes $$y+3=\pm \frac{3}{2}(x-2)$$, and foci at $$\displaystyle \left(2 \pm \frac{\sqrt{455}}{6}, -3\right)$$.
The equation represents a hyperbola centered at (2, -3) with a horizontal transverse axis. Its vertices are at $$\left(2 \pm \frac{\sqrt{35}}{3}, -3\right)$$, asymptotes $$y+3=\pm \frac{3}{2}(x-2)$$, and foci at $$\left(2 \pm \frac{\sqrt{455}}{6}, -3\right)$$.

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