Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$25 x^{2}-9 y^{2}+100 x-54 y+10=0$$

Step-by-Step Solution

Answer

Vertices at (1, -3) and (-5, -3); foci at (1.83, -3) and (-7.83, -3); asymptotes: y = (5/3)x + 16/3, y = -(5/3)x - 4/3.

Step 1: Rearrange and Group the Variables

Start by rearranging the equation with respect to the terms involving \(x\) and \(y\):\[ 25x^2 + 100x - 9y^2 - 54y + 10 = 0 \]Group the terms involving \(x\) and \(y\): \[(25x^2 + 100x) - (9y^2 + 54y) = -10 \]

Step 2: Complete the Square for x

Complete the square for the \(x\) terms:\[ 25(x^2 + 4x) = 25((x+2)^2 - 4) = 25(x+2)^2 - 100 \]

Step 3: Complete the Square for y

Complete the square for the \(y\) terms:\[ -9(y^2 + 6y) = -9((y+3)^2 - 9) = -9(y+3)^2 + 81 \]

Step 4: Simplify the Equation

Substitute both completed squares back into the equation:\[ 25(x+2)^2 - 100 - 9(y+3)^2 + 81 = -10 \]\[ 25(x+2)^2 - 9(y+3)^2 - 19 = -10 \] This simplifies to: \[ 25(x+2)^2 - 9(y+3)^2 = 9 \]

Step 5: Divide Throughout to Form Hyperbola Standard Equation

Divide the entire equation by 225 to get the hyperbola's standard form:\[ \frac{(x+2)^2}{9} - \frac{(y+3)^2}{25} = 1 \] This equation now represents a hyperbola centered at \((-2, -3)\).

Step 6: Identify Vertices and Foci

For the hyperbola \( \frac{(x+2)^2}{9} - \frac{(y+3)^2}{25} = 1 \):- The vertices are at \((-2 \pm 3, -3)\). Thus, the vertices are \((1, -3)\) and \((-5, -3)\).- Compute \(c\) for the foci using \(c^2 = a^2 + b^2\): \[ c^2 = 9 + 25 = 34 \] \[ c = \sqrt{34} \approx 5.83 \]- The foci are at \((-2 \pm \sqrt{34}, -3)\). Thus, the foci are approximately \((1.83, -3)\) and \((-7.83, -3)\).

Step 7: Find the Equations of Asymptotes

The asymptotes have the equation in the form:\[ y = k \pm \frac{b}{a}(x - h) \]Substituting the values, we get the asymptote equations:\[ y + 3 = \pm \frac{5}{3}(x + 2) \]Simplifying, the asymptotes are:\[ y = \frac{5}{3}x + \frac{16}{3} \]\[ y = -\frac{5}{3}x - \frac{4}{3} \]

Step 8: Sketch the Graph

Draw the hyperbola using the center \((-2,-3)\), vertices, and asymptotes as guides:- Plot \( (-2,-3) \) as the center.- Plot the vertices:\( (1,-3) \) and \((-5,-3)\).- Draw the asymptotes: \(y = \frac{5}{3}x + \frac{16}{3}\) and \(y = -\frac{5}{3}x - \frac{4}{3}\). Connect these to form the hyperbola.

Key Concepts

Vertices of Hyperbola

In a hyperbola, vertices are crucial points that mark the extremities of the central axis. They give your hyperbola its defining shape. For a hyperbola given in the standard form: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]its vertices depend directly on the value of \(a\). Here, the problem reduced to \[ \frac{(x+2)^2}{9} - \frac{(y+3)^2}{25} = 1 \]Hence, the center is \((-2, -3)\), and \(a^2 = 9\). This tells us \(a = 3\).

• The vertices are situated along the x-axis, offset by \(\pm a\) from the center.
• So, vertices will be: \((-2 + 3, -3) = (1, -3)\) and \((-2 - 3, -3) = (-5, -3)\).

These are the points where the hyperbola starts curving away from its center.

Foci of Hyperbola

The foci of a hyperbola are integral to its geometric definition, playing a part in the set of points making the hyperbola's shape. The foci lie further along the axis than the vertices. They are calculated using:\[ c^2 = a^2 + b^2 \]For our hyperbola, we find \( c^2 = 9 + 25 = 34 \), and \( c = \sqrt{34} \), approximately \(5.83\).

• The foci, offset from the center \((-2, -3)\), become: \((-2 \pm c, -3)\).
• Numerically, this is \((1.83, -3)\) and \((-7.83, -3)\).

These points indicate where the hyperbola's branches direct towards.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are lines that the curve approaches as it stretches towards infinity. They give the hyperbola its open and endless nature. The equations of these lines for the hyperbola:\[ \frac{(x+2)^2}{9} - \frac{(y+3)^2}{25} = 1 \]are obtained through:\[ y = k \pm \frac{b}{a}(x - h) \]where \(a = 3, b = 5, h = -2, k = -3\). Thus,

• The slope \(\frac{b}{a}\) is \(\frac{5}{3}\).
• Plug these into the formula to get:
  • \( y + 3 = \frac{5}{3}(x + 2) \)
  • \( y + 3 = -\frac{5}{3}(x + 2) \)

After rearranging:\[ y = \frac{5}{3}x + \frac{16}{3} \] and \[ y = -\frac{5}{3}x - \frac{4}{3} \].These lines guide the direction in which the hyperbola stretches.

Completing the Square

Completing the square is an algebraic technique used to transform quadratic equations into a perfect square trinomial. This process is key to simplifying the equation of a hyperbola into its standard form. For our equation, \[ 25x^2 + 100x - 9y^2 - 54y + 10 = 0 \], it involves:

• Grouping the \(x\) and \(y\) terms separately: \[ (25x^2 + 100x) - (9y^2 + 54y) = -10 \]
• Completing the square for \(x\) terms: \(25(x+2)^2 - 100\).
• Completing the square for \(y\) terms: \(-9(y+3)^2 + 81\).

Insert these back to get \[ 25(x+2)^2 - 9(y+3)^2 = 9 \]. This reshaped equation centers the hyperbola and prepares it for graph interpretation.

Graphing Hyperbolas

Graphing hyperbolas once you have their equation in standard form is much simpler. It provides a visual representation of how all elements such as vertices, foci, and asymptotes interact in the coordinate plane. Begin with the simplified equation, \[ \frac{(x+2)^2}{9} - \frac{(y+3)^2}{25} = 1 \].

• Identify the center: \((-2, -3)\). Mark this point.
• Plot the vertices: \((1, -3)\) and \((-5, -3)\).
• Draw asymptotes to establish the direction and spread of the hyperbola.
  • \(y = \frac{5}{3}x + \frac{16}{3}\)
  • \(y = -\frac{5}{3}x - \frac{4}{3}\)
• Sketch the hyperbola so it approaches but never touches the asymptotes.

The visual graph brings mathematics to life and solidifies the hyperbola's structure.