Discuss the conic section $$ 73 x^{2}+72 x y+52 y^{2}+30 x-40 y-75=0 $$

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1. Define the quadratic coefficients from
   $$
   73 x^{2}+72 x y+52 y^{2}+30 x-40 y-75=0:
   $$
   $$
   A=73,\quad B=72,\quad C=52,\quad D=30,\quad E=-40,\quad F=-75.
   $$

2. Compute the discriminant to classify the conic. The discriminant is
   $$
   \Delta = B^2 - 4AC = 72^2 - 4\cdot73\cdot52.
   $$
   Calculate:
   $$
   72^2 = 5184,\qquad 4\cdot73\cdot52 = 4\cdot3796 = 15184.
   $$
   Therefore,
   $$
   \Delta = 5184 - 15184 = -10000.
   $$
   Since $$\Delta < 0$$, the conic is an ellipse (provided it is nondegenerate).

3. Because the equation contains an $$xy$$ term, the ellipse is rotated. The angle of rotation $$\theta$$ is given by
   $$
   \tan(2\theta) = \frac{B}{A-C}.
   $$
   Compute:
   $$
   A - C = 73 - 52 = 21,\quad \tan(2\theta)=\frac{72}{21}=\frac{24}{7}.
   $$
   Thus,
   $$
   \theta=\frac{1}{2}\arctan\left(\frac{24}{7}\right).
   $$

4. Find the center of the ellipse. The center $$(x_0,y_0)$$ is found by solving
   $$
   \frac{\partial}{\partial x}(73x^2+72xy+52y^2+30x-40y-75)=0,
   $$
   $$
   \frac{\partial}{\partial y}(73x^2+72xy+52y^2+30x-40y-75)=0.
   $$
   This yields the system:
   $$
   146x + 72y + 30 = 0,\quad (1)
   $$
   $$
   72x + 104y - 40 = 0.\quad (2)
   $$
   Solve Equation (1) for $$x$$:
   $$
   146x = -30 - 72y \quad\Longrightarrow\quad x = \frac{-30-72y}{146}.
   $$
   Substitute into Equation (2):
   $$
   72\left(\frac{-30-72y}{146}\right) + 104y = 40.
   $$
   Multiply both sides by 146 to eliminate the denominator:
   $$
   -2160 - 5184y + 15184y = 5840.
   $$
   Combine like terms:
   $$
   10000y - 2160 = 5840 \quad\Longrightarrow\quad 10000y = 8000.
   $$
   Hence,
   $$
   y = 0.8.
   $$
   Substitute back to get $$x$$:
   $$
   x = \frac{-30 - 72(0.8)}{146} = \frac{-30 - 57.6}{146} = \frac{-87.6}{146} \approx -0.6.
   $$
   So, the center is
   $$
   (x_0,y_0) \approx (-0.6,\,0.8).
   $$

5. Eliminate the mixed term by diagonalizing the quadratic form. The quadratic part can be written in matrix form using
   $$
   \begin{pmatrix} x & y \end{pmatrix}
   \begin{pmatrix} 73 & 36 \ 36 & 52 \end{pmatrix}
   \begin{pmatrix} x \ y \end{pmatrix},
   $$
   where the coefficient $$36$$ comes from halving $$B=72$$. Find the eigenvalues $$\lambda$$ by solving
   $$
   \det\begin{pmatrix}73-\lambda & 36 \ 36 & 52-\lambda \end{pmatrix}=0.
   $$
   Compute:
   $$
   (73-\lambda)(52-\lambda)-36^2=0.
   $$
   Expand and simplify:
   $$
   \lambda^2 - (73+52)\lambda + (73\cdot52-1296)=\lambda^2-125\lambda+2500=0.
   $$
   The discriminant of this quadratic is
   $$
   125^2-4(2500)=15625-10000=5625,
   $$
   hence
   $$
   \sqrt{5625} = 75.
   $$
   The eigenvalues are
   $$
   \lambda_1=\frac{125-75}{2}=25,\qquad \lambda_2=\frac{125+75}{2}=100.
   $$

6. After translating to the center and rotating to align with the principal axes, the equation takes the form
   $$
   25X'^2 + 100Y'^2 = 100,
   $$
   where $$(X',Y')$$ are the coordinates in the rotated frame. Dividing both sides by 100 gives
   $$
   \frac{X'^2}{4} + \frac{Y'^2}{1} = 1.
   $$
   This is the standard form of an ellipse with
   $$
   a^2 = 4,\quad b^2 = 1,
   $$
   so that
   $$
   a = 2,\quad b = 1.
   $$
   The major axis corresponds to the eigenvalue $$25$$ (since a larger denominator gives a larger axis length) and the minor axis corresponds to the eigenvalue $$100$$.

7. In summary, the given conic section is a rotated ellipse with:
   - Center at $$\left(-0.6,\,0.8\right)$$,
   - Rotation angle $$\theta = \frac{1}{2}\arctan\left(\frac{24}{7}\right)$$,
   - Semi-major axis length $$2$$ and semi-minor axis length $$1$$.

Thus, the conic section
$$
73 x^{2}+72 x y+52 y^{2}+30 x-40 y-75=0
$$
represents a rotated ellipse as described above.
The given equation represents a rotated ellipse with center at approximately $$(-0.6, 0.8)$$, a semi-major axis length of 2, and a semi-minor axis length of 1.

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