Problem 36
Question
Rewrite the equation \(4 x^{2}-9 y^{2}=36\) in standard form. Then write the equation for a translation right 3 units and down 5 units. Draw the graph of each.
Step-by-Step Solution
Verified Answer
The equation in standard form is \((x/3)^{2} - (y/2)^{2} = 1\). After translation, the new equation is \(((x-3)/3)^{2} - ((y+5)/2)^{2} = 1\).
1Step 1: Convert equation into standard form
Divide every term in the equation \(4 x^{2}-9 y^{2}=36\) by 36 to get the equation in standard form. This will yield \((x/3)^{2} - (y/2)^{2} = 1\).
2Step 2: Interpret the standard form equation
The given equation is of a hyperbola because the equation takes the form \((x/h)^{2} - (y/k)^{2} = 1\). The graph opens left and right. The center of the hyperbola is at the origin (0,0).
3Step 3: Translate the equation
The equation for a translation right 3 units and down 5 units is given by \(((x-3)/3)^{2} - ((y+5)/2)^{2} = 1\), which is obtained by subtracting each translation unit from the corresponding variable.
Key Concepts
Standard Form of HyperbolaTranslation of GraphsCoordinate Geometry
Standard Form of Hyperbola
The standard form of a hyperbola is crucial for understanding its structure. It looks like this:
In the original exercise, determining the standard form began by dividing every term of the equation \( 4x^2 - 9y^2 = 36 \) by 36. This simplification transforms it to \( \left(\frac{x}{3}\right)^2 - \left(\frac{y}{2}\right)^2 = 1 \). This form reveals a horizontally-opening hyperbola centered at the origin (0,0). Hence, the roles of \( h \) and \( k \) are played by 0, since there are no translations involved yet.
Understanding these components allows for further operations like graph plotting and translations, as explored in other sections.
- \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
- \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \)
In the original exercise, determining the standard form began by dividing every term of the equation \( 4x^2 - 9y^2 = 36 \) by 36. This simplification transforms it to \( \left(\frac{x}{3}\right)^2 - \left(\frac{y}{2}\right)^2 = 1 \). This form reveals a horizontally-opening hyperbola centered at the origin (0,0). Hence, the roles of \( h \) and \( k \) are played by 0, since there are no translations involved yet.
Understanding these components allows for further operations like graph plotting and translations, as explored in other sections.
Translation of Graphs
Translating graphs is a useful technique in coordinate geometry. It involves shifting the entire graph of a function horizontally and/or vertically. This kind of translation doesn’t alter the graph’s shape, but it does change its position on the coordinate plane.
For the given problem, translating the hyperbola involves shifting it to the right by 3 units and down by 5 units. This changes the center from (0,0) to (3,-5).
To translate a hyperbola, modify the standard form by updating the values of \( h \) and \( k \). For a translation to the right by 3 units, decrease \( h \) by 3 (making it \( x-3 \)), and for a downward shift by 5 units, increase \( k \) by 5 (making it \( y+5 \)).
The newly transformed equation therefore becomes \( \left( \frac{x-3}{3} \right)^2 - \left( \frac{y+5}{2} \right)^2 = 1 \). Here, the changes in the equations aptly reflect the shifts in position, while retaining the structure of a horizontally-opening hyperbola.
For the given problem, translating the hyperbola involves shifting it to the right by 3 units and down by 5 units. This changes the center from (0,0) to (3,-5).
To translate a hyperbola, modify the standard form by updating the values of \( h \) and \( k \). For a translation to the right by 3 units, decrease \( h \) by 3 (making it \( x-3 \)), and for a downward shift by 5 units, increase \( k \) by 5 (making it \( y+5 \)).
The newly transformed equation therefore becomes \( \left( \frac{x-3}{3} \right)^2 - \left( \frac{y+5}{2} \right)^2 = 1 \). Here, the changes in the equations aptly reflect the shifts in position, while retaining the structure of a horizontally-opening hyperbola.
Coordinate Geometry
Coordinate geometry, or analytic geometry, allows the study of geometry using a coordinate system. It's an essential part of understanding hyperbolas because it facilitates precise graph plotting.
The beauty of coordinate geometry lies in the seamless way it blends algebra with geometry. By using coordinates, complex geometric problems are resolved using algebraic equations.
For hyperbolas, coordinate geometry helps in plotting points, defining the axes, and evaluating curves. The equation in the standard form lets you plot points around the center, depending on the values of \( a \) and \( b \) derived from the equation. The center's coordinates (\( h, k \)) aid in graphing accurately, whether at the origin or translated elsewhere.
This mathematical language not only assists in plotting hyperbolas but also in understanding their asymptotes, foci, and lengths between vertices. Thus, prying deeper into the algebra of a hyperbola unveils broader geometric truths, empowering further learning and application in diverse mathematical branches.
The beauty of coordinate geometry lies in the seamless way it blends algebra with geometry. By using coordinates, complex geometric problems are resolved using algebraic equations.
For hyperbolas, coordinate geometry helps in plotting points, defining the axes, and evaluating curves. The equation in the standard form lets you plot points around the center, depending on the values of \( a \) and \( b \) derived from the equation. The center's coordinates (\( h, k \)) aid in graphing accurately, whether at the origin or translated elsewhere.
This mathematical language not only assists in plotting hyperbolas but also in understanding their asymptotes, foci, and lengths between vertices. Thus, prying deeper into the algebra of a hyperbola unveils broader geometric truths, empowering further learning and application in diverse mathematical branches.
Other exercises in this chapter
Problem 35
Write the equation of the circle that passes through the given point and has a center at the origin. (Hint: You can use the distance formula to find the radius.
View solution Problem 36
Find the foci for each equation of an ellipse. $$ 25 x^{2}+4 y^{2}=100 $$
View solution Problem 36
Graph each circle so that the center is at the origin. Then write the equation. radius \(\frac{1}{2}\)
View solution Problem 36
Write an equation of a parabola with a vertex at the origin. \(\operatorname{directrix} x=-3\)
View solution