Problem 44 - 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. $$ 4 x^{2}-y^{2}+32 x+6 y+39=0 $$

Step-by-Step Solution

Answer

The standard form of the given equation is \(\frac{(x + 4)^{2}}{4} - \frac{(y - 3)^{2}}{16} = 1 \). The foci are located at \((-4\pm\sqrt{20}, 3)\). The equations of the asymptotes are given by \(y = 3 \pm 2(x + 4)\).

Step 1: Rearranging Terms

First step is to rearrange the given terms to clear any constants and to group like terms. We achieve this by moving the constant term, 39, to the other side: \(4x^{2} + 32x - y^{2} + 6y = -39\).

Step 2: Completing the Square for x and y

Now, we perform completing the square for both x and y terms. We do this by halving the coefficient of x and y on the squared term, squaring it and adding to both sides. We have \(4(x^{2} + 8x + 16) - (y^{2} - 6y + 9) = -39 + 64 - 9\).

Step 3: Simplifying the Equation

Upon simplifying, we find \(4(x + 4)^{2} - (y - 3)^{2} = 16\). Divide throughout by 16 to achieve standard form \(\frac{(x + 4)^{2}}{4} - \frac{(y - 3)^{2}}{16} = 1 \).

Step 4: Determine the Asymptotes and Foci

The equation of asymptotes of a hyperbola in standard form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1 \) is given by \(y = k \pm \frac {b}{a} (x-h) \). Thus, our equations will be \(y = 3 \pm 2(x + 4)\). To find the foci, we calculate \(c = \sqrt{a^2 + b^2}\), which gives \(c = \sqrt{4+16} =\sqrt{20} \). Thus, our foci would be at \((-4\pm\sqrt{20}, 3)\)

Step 5: Graphing the Hyperbola

Finally, we sketch the graph for the standard equation. We plot the center at (-4, 3), plot a rectangle using the values of a and b as lengths of the sides, sketch the asymptotes using the rectangle diagonals, and draw the two branches of the hyperbola touching the rectangle at the vertices.

Key Concepts

Completing the Square

In algebra, completing the square is a powerful technique used to transform quadratic expressions into a form that reveals the roots or vertex of the function. It is especially useful in solving equations and graphing conic sections like hyperbolas. To complete the square, one must form a perfect square trinomial from a quadratic expression, which typically involves the following steps:

• Group the variable terms and move the constant to the opposite side of the equation.
• For each variable, take half of the coefficient of the linear term, square it, and add it to both sides of the equation.
• Factor the perfect square trinomial on one side and simplify the other side.

By using this method, the given hyperbola equation transforms from its general form into a more workable version that sets the stage for finding the standard form of the hyperbola.

Standard Form of a Hyperbola

The standard form of a hyperbola is crucial for understanding its structure and characteristics, such as its orientation, center, vertices, and axes. For a hyperbola centered at point \(h, k\), the standard forms are:
\[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\] for a horizontal hyperbola, and
\[\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\] for a vertical hyperbola.
Here, \(a\) and \(b\) are distances that relate to the lengths of the axes of symmetry and help to locate the vertices. The transformation to standard form allows you to graph the hyperbola accurately and provides the foundation for determining the positions of significant features such as the asymptotes and foci.

Hyperbola Asymptotes

Asymptotes are imaginary lines that the hyperbola approaches but never touches. They act as guides for sketching the hyperbola and are important geometric references. The equation of asymptotes for any hyperbola in standard form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\) or \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\) is:
\[y = k \pm \frac{b}{a}(x - h)\]
In the exercise, the standard form equation is transformed into the necessary shape to identify the equation of the asymptotes. Knowing the asymptotes' equations is essential for drawing an accurate sketch of the hyperbola.

Hyperbola Foci

The foci (plural of focus) of a hyperbola are key to its definition and are points located along the axis of symmetry from which the hyperbola is derived. Their position is dependent on the standard form parameters \(a\) and \(b\). To find the foci of a hyperbola with equation \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), you calculate the distance \(c\) from the center using the relation \(c^2 = a^2 + b^2\). The coordinates of the foci are then given as \((h \pm c, k)\) or \((h, k \pm c)\) depending on the hyperbola's orientation. The foci are integral to the hyperbola’s property that for any point on the hyperbola, the difference in distances to the foci is constant.