1. **Problem 1: Find the general equation of the hyperbola with vertices (0, ±2) and foci (0, ±2\sqrt{5})**. Step 1: Identify the orientation and parameters. - The vertices are on the y-axis at (0, ±2), so the hyperbola is vertical. - Center is at the origin (0,0). - Distance from center to vertex: $a=2$. - Distance from center to foci: $c=2\sqrt{5}$. Step 2: Use relationship between $a$, $b$, and $c$ for hyperbolas: $$ c^2 = a^2 + b^2 $$ Substitute values: $$ (2\sqrt{5})^2 = 2^2 + b^2 $$ $$ 4 \times 5 = 4 + b^2 $$ $$ 20 = 4 + b^2 $$ $$ b^2 = 16 $$ Step 3: Write the standard form equation for vertical hyperbola centered at origin: $$ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $$ Substitute values: $$ \frac{y^2}{4} - \frac{x^2}{16} = 1 $$ --- 2. **Problem 2: Find an equation of the hyperbola in standard form having a focus at (5, 0) and ends of its conjugate axis at (0, ±2)**. Step 1: Analyze given points. - Focus at (5,0) means hyperbola is horizontal and center is on x-axis. - Ends of conjugate axis at (0, ±2) means center y=0 and conjugate axis length is vertical. Step 2: The conjugate axis length is $2b=4$, so $b=2$. - Since conjugate axis endpoints are at (0, ±2), center at (0,0). Step 3: The focus is at (5,0), so $c=5$ (distance from center). Step 4: Use relationship: $$ c^2 = a^2 + b^2 $$ $$ 5^2 = a^2 + 2^2 $$ $$ 25 = a^2 + 4 $$ $$ a^2 = 21 $$ Step 5: Write the standard form for horizontal hyperbola centered at origin: $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$ Substitute: $$ \frac{x^2}{21} - \frac{y^2}{4} = 1 $$ --- **Final answers:** 1) $$ \frac{y^2}{4} - \frac{x^2}{16} = 1 $$ 2) $$ \frac{x^2}{21} - \frac{y^2}{4} = 1 $$