Problem 49 - Algebra and Trigonometry

Question

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. $$ 4 x^{2}-25 y^{2}-32 x+164=0 $$

Step-by-Step Solution

Answer

The standard form of hyperbola is \((x-4)^{2}/9 - y^{2}/1.44 = 1\), the graph will have vertices at (7,0) and (1,0), co-vertices at (4,±1.2) and the foci at approximately (7.23,0) and (0.78,0). The equations of asymptotes are \(y=\pm (1.2/3)(x-4)\).

Step 1: Rearranging the Terms

First, we should group the x's and y's terms together and move the number to the right side of the equation. So we rewrite the equation as: \( 4x^{2}-32x - 25y^{2} = -164 \)

Step 2: Completing the square

We complete the square separately for the x’s and y’s and then organize the equation with the constant on the right side. In the x's terms, the coefficient of \(x^{2}\) is 4 and the coefficient of x is -32. For completing the square, we take half of the coefficient of x, square it and add it to both sides of the equation. This gives us: \( 4(x^{2}-8x+16) - 25y^{2} = -164 + 4*16 \) which simplifies to \( 4(x-4)^{2} - 25y^{2} = -36 \)

Step 3: Converting to standard form

Now, we will convert the equation to standard form. For this, we'll isolate y by moving all of the other terms to the right side. After doing this, we divide all terms by -36 to get the equation to standard form which can be written as \( (x-4)^{2}/9 - y^{2}/1.44 = 1\)

Step 4: Graphing the Hyperbola

Now that we have the standard equation of the hyperbola. We can sketch it. The centre is at (4,0). The lengths of semi-major and semi-minor axis are \(\sqrt{9}=3\) and \(\sqrt{1.44}\) respectively. So, vertices are at (4±3, 0) and co-vertices at (4,±\(\sqrt{1.44}\)). Draw the box and diagonals to plot the asymptotes and the hyperbola.

Step 5: The Foci and Asymptotes

To find the foci, we use the formula \(c =\sqrt{a^{2} + b^{2}}\), where a and b are the lengths of semi-major and semi-minor axis respectively. This gives us √(9 + 1.44) = √10.44 = about 3.23. Meaning the foci are at coordinates (4±3.23, 0). The equations of the asymptotes are given by \(y=\pm (b/a)(x-h)+k\), where (h,k) are the coordinates of the center of the hyperbola, a is the length of semi-major axis and b is the length of semi-minor axis and the energy sign refers to the upward and downward slope of the asymptotes. Substituting the given values we get, equations of the asymptotes as \(y=\pm (\sqrt{1.44}/3)(x-4)\).

Key Concepts

Completing the Square

The process of completing the square involves transforming a quadratic equation into a perfect square trinomial. This technique simplifies the equation and is particularly useful in converting equations into their standard forms.

In the provided exercise, we first rearrange the terms of the equation to group the x and y variables separately. For the x terms specifically, the equation is manipulated to look like a completed square.

Here's how it works: take the coefficient of the linear x term, which in our case is -32, divide it by 2 to get -16, and then square it to obtain 256.

• Add this squared value to both sides, ensuring balance in the equation.
• The expression becomes a perfect square trinomial, \(4(x-4)^2\), after factoring.

Completing the square in this manner simplifies equations and prepares them for conversion to standard form.

Standard Form

Standard form represents the hyperbola equation in a format that is more understandable and useful for graphing. The general format for the standard form of a hyperbola is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).

With our hyperbola, after completing the square and rearranging, the terms are divided by the constant on the right to attain this form. We divide through by -36 to normalize it.

• This results in \((x-4)^2/9 - y^2/1.44 = 1\)

Here, \(a^2 = 9\) and \(b^2 = 1.44\). The center of the hyperbola (h, k) has now been revealed as (4, 0).

This form makes further calculations, like locating foci and asymptotes, straightforward.

Foci

Foci are crucial points inside a hyperbola along its major axis, and they help define its shape. They are located by using the relationship \(c = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are derived from the standard form.

In our example, \(a^2 = 9\) and \(b^2 = 1.44\), hence the distance to the foci from the center is calculated as:

• \(c = \sqrt{9 + 1.44} = \sqrt{10.44} ≈ 3.23\)

With the center at (4,0), the foci are positioned horizontally at coordinates \(4 ± 3.23, 0\).

These points are important when sketching the hyperbola as they determine its orientation and curvature.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never touches. They are vital for graphing because they establish the directions and confines of the curve.

The equations for asymptotes of a hyperbola centered at \(h, k\) are \(y = \pm \frac{b}{a}(x-h) + k\).

• For our hyperbola, inserting the necessary values gives us the equations: \(y = \pm (\sqrt{1.44}/3)(x-4)\).

These indicate the slope of the lines relative to the center, providing a guide for sketching the hyperbola accurately.

Knowing the asymptotes allows us to predict how sharply the hyperbola opens and visualize the complete graph effectively.