In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $4x^2-25y^2-32x+164=0$

Verified step by step guidance

1. Start with the given equation: \$4x^{2} - 25y^{2} - 32x + 164 = 0$.
2. Group the \(x\) terms and the \(y\) terms separately: \$4x^{2} - 32x - 25y^{2} + 164 = 0$.
3. Factor out the coefficient of \(x^{2}\) from the \(x\) terms: \$4(x^{2} - 8x) - 25y^{2} + 164 = 0$.
4. Complete the square for the expression inside the parentheses for \(x\): take half of \(-8\), which is \(-4\), square it to get \$16\(, then add and subtract \)16\( inside the parentheses: \)4(x^{2} - 8x + 16 - 16) - 25y^{2} + 164 = 0$.
5. Rewrite the equation using the perfect square trinomial and simplify constants: \$4(x - 4)^{2} - 25y^{2} + 164 - 4 imes 16 = 0$, then move constants to the other side to isolate the conic.

Key Concepts

Completing the Square

Completing the square is a method used to rewrite quadratic expressions in the form (x - h)² or (y - k)² by adding and subtracting terms. This technique helps convert the given equation into a standard form, making it easier to identify the center, vertices, and other properties of conic sections like hyperbolas.

Standard Form of a Hyperbola

The standard form of a hyperbola is (x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/a² - (x - h)²/b² = 1, where (h, k) is the center. Converting the equation into this form reveals key features such as the center, vertices, and orientation, which are essential for graphing and further analysis.

Foci and Asymptotes of a Hyperbola

The foci are two fixed points inside the hyperbola that define its shape, located at a distance c from the center, where c² = a² + b². Asymptotes are lines that the hyperbola approaches but never touches, given by equations derived from the standard form. Identifying these helps in accurately sketching the hyperbola.