1. The problem involves identifying and analyzing the conic sections given by equations of ellipses, hyperbolas, and second-degree polynomials in $x$ and $y$. 2. For each explicit ellipse equation, they are given in the form $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$ where $(h,k)$ is the center and $a^2, b^2$ are the denominators representing the squares of axis lengths. 3. For the hyperbola, the form is $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,$$ indicating the transverse axis is along the $x$-axis. 4. For the general quadratic forms, we complete the square to rewrite them in standard conic form. --- ### Detailed Analysis of Each Equation: **Equation 1:** Ellipse at top-left $$(x-22)^2 / 35 + (y+12)^2 / 65 = 1$$ - Center: $(22,-12)$ - Semi-major and semi-minor axes are $\sqrt{35}$ and $\sqrt{65}$ **Equation 2:** Hyperbola at top-right $$(x+4)^2 / 40 - (y-3)^2 / 26 = 1$$ - Center: $(-4,3)$ - Transverse axis length $2\sqrt{40}$ along $x$ - Conjugate axis length $2\sqrt{26}$ along $y$ **Equation 3:** Ellipse at center $$(x+2)^2 / 36 + (y-3)^2 / 64 = 1$$ - Center: $(-2,3)$ - Axes: 6 (since $\sqrt{36}=6$) and 8 **Equation 4:** General quadratic $$9x^2 + 25y^2 - 36x + 50y - 164 = 0$$ - Complete the square: $$9x^2 - 36x + 25y^2 + 50y = 164$$ Divide each group: $$9(x^2 - 4x) + 25(y^2 + 2y) = 164$$ Complete squares: $$9[(x^2 - 4x + 4) - 4] + 25[(y^2 + 2y + 1) - 1] = 164$$ $$9(x - 2)^2 - 36 + 25(y + 1)^2 - 25 = 164$$ $$9(x-2)^2 + 25(y+1)^2 = 164 + 36 + 25 = 225$$ Divide through by 225: $$\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1$$ - Ellipse centered at $(2,-1)$ with semi-axes 5 and 3 **Equation 5:** Ellipse at bottom-left $$(x+2)^2 / 15 + (y-2)^2 / 20 = 1$$ - Center $(-2,2)$ **Equation 6:** General quadratic $$9y^2 - 16x^2 + 32x + 144y - 16 = 0$$ Rewrite grouping $x$ and $y$: $$-16x^2 + 32x + 9y^2 + 144y = 16$$ Divide coefficients: $$-16(x^2 - 2x) + 9(y^2 + 16y) = 16$$ Complete the square: $$-16[(x^2 - 2x + 1) - 1] + 9[(y^2 + 16y + 64) - 64] = 16$$ $$-16(x-1)^2 + 16 + 9(y+8)^2 - 576 = 16$$ $$-16(x-1)^2 + 9(y+8)^2 - 560 = 16$$ $$-16(x-1)^2 + 9(y+8)^2 = 576$$ Divide through by 576: $$-\frac{(x-1)^2}{36} + \frac{(y+8)^2}{64} = 1$$ Rewrite to standard hyperbola form: $$\frac{(y+8)^2}{64} - \frac{(x-1)^2}{36} = 1$$ - Hyperbola centered at $(1,-8)$ with vertical transverse axis **Equation 7:** Ellipse at bottom-right $$(x-5)^2 / 17 + (y-3)^2 / 27 = 1$$ - Center $(5,3)$ --- ### Summary: - Equations 1,3,4,5,7 are ellipses; - Equations 2 and 6 are hyperbolas. ### Desmos field (minimal): - No explicit function $y = f(x)$ given, so just leave minimal. Final answer for standard forms and centers calculated above.