Problem 37 - Algebra and Trigonometry

Question

\(35-38\) Use a graphing device to graph the conic. $$ 9 x^{2}+36=y^{2}+36 x+6 y $$

Step-by-Step Solution

Answer

The conic is a hyperbola centered at (2, -3) with a horizontal transverse axis.

Step 1: Simplify the Equation

Start by arranging the given equation: \(9x^2 + 36 = y^2 + 36x + 6y\). Move all terms to one side to get it in standard conic form. Subtract \(y^2 + 36x + 6y\) from both sides to get: \(9x^2 - y^2 - 36x - 6y + 36 = 0\).

Step 2: Complete the Square for x

Focus on the terms involving \(x\): \(9x^2 - 36x\). Factor out the coefficient of \(x^2\): \(9(x^2 - 4x)\). Complete the square: take half of \(4\), square it to get \(4\), and add/subtract inside the parenthesis to get \(9(x^2 - 4x + 4 - 4)\) which simplifies to \(9((x-2)^2 - 4)\).

Step 3: Complete the Square for y

Focus on the terms involving \(y\): \(-y^2 - 6y\). Factor out \(-1\): \(-(y^2 + 6y)\). Complete the square: take half of \(6\), square it to get \(9\), and add/subtract inside to get \(-(y^2 + 6y + 9 - 9)\) which simplifies to \(-(y+3)^2 + 9\).

Step 4: Rewrite the Equation

Substitute back the completed squares into the equation: \(9((x-2)^2 - 4) - ((y+3)^2 - 9) + 36 = 0\). Simplify this to get \(9(x-2)^2 - (y+3)^2 - 36 + 9 + 36 = 0\).

Step 5: Confirm the Hyperbola Form

The equation simplifies to \(9(x-2)^2 - (y+3)^2 = 9\). Divide the entire equation by \(9\) to normalize: \((x-2)^2 - \frac{(y+3)^2}{9} = 1\). This is a standard form of a hyperbola centered at \((2, -3)\).

Step 6: Graph with a Graphing Device

Use a graphing calculator or software to enter the equation \((x-2)^2 - \frac{(y+3)^2}{9} = 1\). This confirms the graph is a hyperbola with a horizontal transverse axis.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that you can think of as an open curve, shaped somewhat like a pair of infinite bows. When graphing a hyperbola, you'll typically notice two separate branches. These branches mirror each other across a center point. Understanding hyperbolas helps in various fields including astronomy and navigation.

Hyperbolas are defined by a set of points where the difference of the distances to two fixed points (called the foci) is constant. Unlike ellipses where this sum is constant, hyperbolas require the difference. The hyperbola has axes – the transverse and conjugate axes. The transverse axis joins the two vertices, and the conjugate axis is perpendicular to it. These axes help define the orientation of the hyperbola.

• Resembles two mirrored curves
• Defined by a constant difference from two focal points
• Two axes: transverse and conjugate

Identifying the type of conic section involved (whether circle, ellipse, parabola, or hyperbola) relies on the general quadratic equation form. This recognition is crucial in solving and graphing conics accurately.

Completing the Square

Completing the square is a mathematical technique used to transform quadratic equations into a form that is easy to analyze and graph. It is often employed in the process of solving quadratic equations and in converting the standard form of conic sections into a form that's more intuitive to graph and understand.

The method involves making a perfect square trinomial from a quadratic expression. This is useful when the expression includes variables squared inside it. Each step in completing the square involves:

• Isolating the quadratic portion
• Taking half of the linear coefficient
• Squaring that result
• Adding and subtracting this square inside the expression

After these steps, terms can be rearranged to reflect a perfect-square polynomial. This manipulation allows variables to express shifts in their positions, effectively translating them if visualized on a graph. Completing the square simplifies equations into easily recognizable forms, such as standard hyperbola, ellipse, or parabola equations.

Graphing Conics

Graphing conic sections like hyperbolas involves transforming a general equation into one that highlights its geometric aspects. Conics include ellipses, parabolas, circles, and hyperbolas, each with unique characteristics and standard equations.

When graphing hyperbolas, recognizing the canonical form is key. This form reveals the center, vertices, and axes of the hyperbola, which are necessary for accurate graph representation. The hyperbola's graph displays:

• Two symmetric branches
• Open ends pointing away from each other
• Defined center and intersecting axes

Using a graphing calculator or software can be immensely helpful. Inputting the standard form accurately displays curves on the graph with their asymptotes – lines the branches approach but never touch. Understanding how to correctly place a hyperbola on a graph enhances comprehension of its geometry and behavior.

Equation of a Hyperbola

The equation of a hyperbola takes two main forms, based on its axis orientation: horizontal or vertical. These forms reflect where the transverse axis lies.

For a hyperbola centered at r(binomial term) and vertex shifts. The general forms are:

• r>Horizontal: \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\) li>
• r>Vertical: \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\) li>

Where \((h, k)\) is the center, and \(a\) and \(b\) are distances from the center to vertices and co-vertices respectively.

Transforming a complex equation into one of these forms often requires completing the square for both variables. This shows how each piece relates to the hyperbola's position and shape. Practicing these steps enables a better grasp of different hyperbola properties, benefitting fields such as physics and engineering where hyperbolic paths occur.