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In Math / Junior High School | 2024-09-08

graph the following equation and discuss briefly the steps. 1. 9x²-16y²-144=0​

Asked by johnreymcatipay

Answer (1)

Step-by-step explanation:To graph the equation \( 9x^2 - 16y^2 - 144 = 0 \), we first need to rewrite it in a standard form. Here are the steps involved:### Step 1: Rearranging the EquationStart by moving the constant term to the other side:\[9x^2 - 16y^2 = 144\]### Step 2: Dividing by 144Next, divide every term by 144 to get the equation in the standard form of a hyperbola:\[\frac{9x^2}{144} - \frac{16y^2}{144} = 1\]This simplifies to:\[\frac{x^2}{16} - \frac{y^2}{9} = 1\]### Step 3: Identify the Standard FormNow we have the equation in the standard form of a hyperbola:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]where \( a^2 = 16 \) and \( b^2 = 9 \). Thus, \( a = 4 \) and \( b = 3 \).### Step 4: Determine the Center and AxesThe center of the hyperbola is at the origin (0,0) because there are no \( h \) or \( k \) terms in the equation. - The transverse axis (which is horizontal since the \( x^2 \) term is positive) has a length of \( 2a = 8 \).- The conjugate axis (vertical) has a length of \( 2b = 6 \).### Step 5: Plotting the AsymptotesThe equations for the asymptotes of this hyperbola are given by:\[y = \pm \frac{b}{a} x = \pm \frac{3}{4} x\]### Step 6: Sketch the Graph1. Begin by plotting the center at (0,0).2. Next, plot points for the vertices along the transverse axis at (4,0) and (-4,0).3. Plot the points for the vertices along the conjugate axis at (0, 3) and (0, -3).4. Draw dashed lines for the asymptotes through the origin, ensuring they extend in both directions.5. Finally, sketch the branches of the hyperbola that open left and right from the center.### ConclusionThe resulting graph represents a hyperbola centered at the origin (0,0) with horizontally opening branches. The vertices are at (4,0) and (-4,0), while the asymptotes guide the shape of the graph. This process allows for a clear understanding of how the hyperbola is oriented and defined by the given equation. If you would like to delve into any specific part or need illustrations, let me know!

Answered by romnickpallon | 2024-09-08