Problem 64 - Algebra 2

Question

What are the coordinates of the foci of \(4 x^{2}-24 x=64-25 y^{2} ?\)

Step-by-Step Solution

Answer

The coordinates of the foci of the given equation are \((3+\sqrt{7}, 0)\) and \((3-\sqrt{7}, 0)\).

Step 1 : Rewrite the equation in standard form

Rewrite the equation as \((x-3)^{2} /16 - y^{2} /9 = 1\), by grouping the x and y terms, completing the square and moving the constant to the other side.

Step 2 : Identify the values of a, b and c

In the rewritten equation, the standard form of an ellipse is \( (x - h)^{2} /a^{2} - (y - k)^{2} /b^{2} = 1.\) In this case, a is the square root of the denominator under the x-term, i.e. \(a = \sqrt{16} = 4\). Similarly, \( b = \sqrt{9} = 3.\) Since \(a^{2} = b^{2} + c^{2}\) for ellipses where the major axis is horizontal, we can find c by rearranging this equation to \(c = \sqrt{a^{2} - b^{2}} = \sqrt{4^{2} - 3^{2}} = \sqrt{7}

Step 3 : Find the coordinates of the foci

Since the x-term in the rewritten equation has a larger denominator, the major axis runs horizontally. Therefore, the coordinates of the foci are given by \((h±c,k)\), which gives us \((3±\sqrt{7},0)\)

Key Concepts

Foci of Hyperbola

The foci of a hyperbola are special points that are essential in understanding its shape. A hyperbola has two branches, and each branch seems to open towards a focus location. Identifying these points helps us describe the hyperbola's orientation accurately.
For a hyperbola written in standard form,

• If the x-term has the larger denominator, the foci lie along the x-axis.
• If the y-term dominates, the foci are aligned along the y-axis.

In the given equation, the x-term's larger denominator signals that the major axis is horizontal, so the foci, \((h \pm c, k)\), indicate horizontal adjustment from the hyperbola's center.

Standard Form of Hyperbola

To solve problems involving hyperbolas, rewriting their equation in standard form is the first crucial step. The standard form helps us determine other important characteristics like foci and axes of symmetry.
A hyperbola's equation has the general form: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] Here:

• \((h, k)\) is the center of the hyperbola.
• \(a^2\) and \(b^2\) are the squares of distances from the center to the vertices and co-vertices.

This makes it much easier to interpret and solve related problems. By reworking the original equation \(4x^2 - 24x = 64 - 25y^2\), we aim to fit this standard mold, revealing a clearer path to understanding the hyperbola's dimensions.

Ellipse Equation Transformation

Though ellipses and hyperbolas differ, transforming an ellipse equation helps in understanding hyperbolas. The process involves similar techniques. Start by isolating terms and solving to fit the forms we know and recognize. Once we approach the target form, it’s about identifying terms that match the pattern.
For hyperbolas:

• Differentiating involves noticing the signs; one term is negative indicating its hyperbolic nature.
• Transforming includes reshaping through algebraic manipulation to follow identifiable standard patterns, such as moving constants and balancing terms.

Transforming the equation brings focus on rewriting as \(\frac{(x-3)^2}{16} - \frac{y^2}{9} = 1\), which clarifies its identity as a hyperbola.

Completing the Square

Completing the square is a method to simplify and solve quadratic equations, used here to convert our equation into that recognizable standard form. It helps in obtaining the center of conic sections like hyperbolas.
With our initial equation: \(4x^2 - 24x = 64 - 25y^2\), completing the square involves:

• Reorganizing the terms into two groups, focusing on the separate handling of x and y terms.
• Adjusting the x's to form a square, which required us to complete it as \( (x - 3)^2 \).

This careful refinement is crucial, allowing us to reframe the equation as a standard hyperbola, leaving the equation easier to understand and analyze.