1. **Problem 9:** Find the equation of an equilateral hyperbola with foci at $(\sqrt{2},0)$ and $(-\sqrt{2},0)$. - An equilateral hyperbola has equal transverse and conjugate axes, so $a = b$. - The foci are at $(\pm c,0)$, where $c^2 = a^2 + b^2$. Since $a = b$, $c^2 = 2a^2$. Given $c = \sqrt{2}$, so $c^2 = 2$. - Thus, $2 = 2a^2 \Rightarrow a^2 = 1$. - The equation of a hyperbola centered at the origin with foci on the x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since $a^2 = b^2 = 1$, the equation is $x^2 - y^2 = 1$. - This matches option A or C (both are the same). 2. **Problem 10:** If eccentricity $e > 1$, what is the conic? - Eccentricity $e$ defines conic types: - $e=0$ circle - $0 < e < 1$ ellipse - $e=1$ parabola - $e > 1$ hyperbola - So the answer is C, hyperbola. 3. **Problem 11:** Identify the graph of $x^2 + 4y^2 = 8x + 8y = 13 = 0$. - The equation seems mistyped; assuming it means $x^2 + 4y^2 - 8x - 8y + 13 = 0$. - Complete the square: - $x^2 - 8x + 4y^2 - 8y = -13$ - $x^2 - 8x + 16 + 4(y^2 - 2y + 1) = -13 + 16 + 4$ - $(x - 4)^2 + 4(y - 1)^2 = 7$ - This is an ellipse centered at $(4,1)$, so answer B. 4. **Problem 12:** Equation of hyperbola with center $(0,0)$, focus $(2,-6)$, vertex $(2,-3)$. - The center is at origin, but focus and vertex are at $(2,-6)$ and $(2,-3)$, so the hyperbola is shifted or the problem has a typo. Assuming the hyperbola is vertical with center at $(2,-4.5)$ (midpoint of vertex and focus). - Distance from center to vertex $a = | -3 + 4.5| = 1.5$, so $a^2 = 2.25$. - Distance from center to focus $c = | -6 + 4.5| = 1.5$, so $c^2 = 2.25$. - Since $c^2 = a^2 + b^2$, $b^2 = c^2 - a^2 = 0$. This is inconsistent, so likely a typo. - Among options, option A $(x - 2)^2/36 - y^2/27 = 1$ fits a hyperbola shifted to $(2,0)$, so choose A. 5. **Eccentricity of ellipse with major axis length 10 and foci $(2,3)$ and $(2,-1)$. - Length of major axis $2a = 10 \Rightarrow a = 5$. - Distance between foci $2c = \sqrt{(2-2)^2 + (3+1)^2} = 4 \Rightarrow c = 2$. - Eccentricity $e = \frac{c}{a} = \frac{2}{5}$. - Answer A. **Final answers:** 9: A 10: C 11: B 12: A Eccentricity: A