Problem 77
Question
Using the same axes, draw the conics \(y=\) \(\pm\left(a x^{2}+1\right)^{1 / 2}\) for \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\) using \(a=\) \(-2,-1,-0.5,-0.1,0,0.1,0.6,1 .\) Make a conjecture about how the shape of the figure depends on \(a\).
Step-by-Step Solution
Verified Answer
The shape changes based on 'a', from inverted (negative 'a') to wider (positive 'a') conics.
1Step 1: Understand the Equation
The equation given is for two conics: \( y = \pm (ax^2 + 1)^{1/2} \). This represents a hyperbola as it contains the term \(x^2\). It will open upwards and downwards, as indicated by the positive and negative square root terms.
2Step 2: Analyze the Influence of 'a'
Examine how different values of \(a\) affect the shape. If \(a = 0\), the equation simplifies to \(y = \pm 1\), a straight line. For negative \(a\), the parabola-like shape is inverted, and for positive \(a\), it becomes wider. Increasing \(a\) or decreasing \(a\) in magnitude affects the 'width' of the curve.
3Step 3: Graph for Each 'a' Value
For each given \(a\): \(-2, -1, -0.5, -0.1, 0, 0.1, 0.6, 1\), substitute into the equation and plot \(y = \pm (ax^2 + 1)^{1/2}\) on the axes within the domain \(-2 \leq x \leq 2\) and range \(-2 \leq y \leq 2\). The curves will appear progressively closer to or further from the \(y\)-axis depending on the value of \(a\).
4Step 4: Interpret the Graphs
For negative \(a\), the curves are sharp and more close (inverted hyperbolas), as \(a\) approaches zero, the hyperbolas widen becoming flatter at \(a=0\). Positive \(a\) values will stretch the hyperbola more, making it wider.
5Step 5: Make a Conjecture
Based on the graphics, we can conjecture that for negative \(a\), the curves are more densely packed and appear inverted, and as \(a\) becomes positive, the curves widen. Thus, the value of \(a\) controls the opening and width of each branch of the conics.
Key Concepts
HyperbolasEffect of ParametersGraphing Equations
Hyperbolas
Hyperbolas are one of the four types of conic sections, which also include circles, ellipses, and parabolas. A hyperbola can be visualized as two symmetric, open curves extending away from each other. In the equation form given here,
- The hyperbola is vertical, meaning it opens upwards and downwards.
- The expressions \( y = \pm (ax^2 + 1)^{1/2} \) define two mirror-image parts of the hyperbola, due to the \( \pm \) sign.
- The squared term \( x^2 \) signifies that the hyperbola branches grow infinitely, as x stretches toward infinity.
Effect of Parameters
Parameters in an equation significantly shape the curve of the graph it represents. Here, the parameter is 'a' found in the hyperbola equation
- When \( a = 0 \), the hyperbola simplifies to the lines \( y = \pm 1 \). This happens because the \( ax^2 \) term disappears, leaving a flat line at \( y = 1 \) and \( y = -1 \).
- For negative values of 'a',
- The hyperbola becomes steeper and inverted, causing the curves to approach each other more closely. This shows a contraction across the x-axis.
- Conversely, when 'a' is positive, the magnitude increase causes the hyperbola to stretch and widen, particularly opening more towards the y-axis.
Graphing Equations
Graphing, in this context, requires plotting the equation \( y = \pm (ax^2 + 1)^{1/2} \) for different 'a' values. Here are the essential steps:
- Select the range: in this exercise, it is specified as \( -2 \leq x \leq 2 \) and \( -2 \leq y \leq 2 \).
- Substitute each value of 'a' into the equation and calculate corresponding y-values for x-values within the given range.
- Plot the points for each resulting equation. \( y \) will take both positive and negative values, representing the two branches of the hyperbola.
Other exercises in this chapter
Problem 75
If a horizontal hyperbola and a vertical hyperbola have the same asymptotes, show that their eccentricities \(e\) and \(E\) satisfy \(e^{-2}+E^{-2}=1\).
View solution Problem 76
Let \(C\) be the curve of intersection of a right circular cylinder and a plane making an angle \(\phi(0
View solution Problem 75
Draw the Folium of Descartes \(x=3 t /\left(t^{3}+1\right)\), \(y=3 t^{2} /\left(t^{3}+1\right) .\) Then determine the values of \(t\) for which this graph is i
View solution