/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Find the points of intersection ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 72

Find the points of intersection of the parabola \(4 x^{2}-8 x=2 y+5\) and the line \(y=15\)

Short Answer

Expert verified
Answer: The points of intersection are \(\left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \(\left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\).

Step by step solution

01

Rewrite the equation of the parabola in terms of \(y\)

We are given the equation of the parabola as \(4 x^2 - 8 x = 2 y + 5\). We can rewrite this equation in terms of \(y\) by solving for \(y\): \(2 y = 4 x^2 - 8 x - 5 \Rightarrow y = 2 x^2 - 4 x - \dfrac{5}{2}\).
02

Set the equations equal to each other

Now we will set the equation of the parabola equal to the equation of the line to find the \(x\)-values of the intersection points: \(2 x^2 - 4 x - \dfrac{5}{2} = 15\).
03

Solve the quadratic equation

Now we need to solve the quadratic equation \(2 x^2 - 4 x - \dfrac{5}{2} - 15 = 0\). First, simplify the equation: \(2 x^2 - 4 x - 35 = 0\). Then, divide the equation by \(2\) to simplify it further: \(x^2 - 2x - 17.5 = 0\). Now, we can use the quadratic formula to solve for the \(x\)-values of the intersection points. The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our case, we have \(a = 1\), \(b = -2\), and \(c = -17.5\). Plugging in these values, we get: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-17.5)}}{2(1)}\). Simplifying further, we find: \(x = \dfrac{2 \pm \sqrt{72}}{2}\). So, the two possible \(x\)-values are: \(x_1 = \dfrac{2 + \sqrt{72}}{2}\) and \(x_2 = \dfrac{2 - \sqrt{72}}{2}\).
04

Find the corresponding \(y\)-values

Now that we have the \(x\)-values of the intersection points, we can find the corresponding \(y\)-values by plugging them back into the equation of the line \(y = 15\). Thus, the coordinates of the intersection points are: \((x_1, y_1) = \left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \((x_2, y_2) = \left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\). So, the points of intersection of the parabola \(4 x^2 - 8 x = 2y + 5\) and the line \(y = 15\) are \(\left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \(\left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are a type of polynomial equation that involves terms up to the second degree (or power). They follow the general form: \[ax^2 + bx + c = 0\]where:\
  • \(a\), \(b\), and \(c\) are constants,\
  • \(x\) is the variable, and\
  • \(a\) cannot be zero.\
In our problem, we had to solve the quadratic equation to find where the parabola intersects the line, leading to:\\[x^2 - 2x - 17.5 = 0\]
To find the solutions, we can utilize the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula gives us potential solutions (also known as the roots) for \(x\). These roots represent the x-values where the curve will intersect the line. Utilizing this method ensures that all potential intersections are considered.
The solutions are real and exist because the discriminant \( b^2 - 4ac \) is positive (72 in this case). When the discriminant is positive, two distinct solutions exist.
Parabola
A parabola is a symmetrical, U-shaped curve that is the graph of a quadratic equation. Parabolas can open either upward or downward, depending on the coefficient of the \(x^2\) term. If this coefficient is positive, the parabola opens upwards; if negative, downwards.
The standard form of a parabola can be represented as:\[y = ax^2 + bx + c\]
In our particular exercise, the equation of the parabola was:\[4x^2 - 8x = 2y + 5\]By rewriting and simplifying, we expressed it as:\[y = 2x^2 - 4x - \frac{5}{2}\]
This equation shows that the parabola opens upwards since the coefficient \(a = 2\) is positive. The vertex, or the highest or lowest point of the parabola, gives insight into its shape and is key to understanding its position relative to the x-axis and other curves like lines.
Line Equations
Lines are described by linear equations. Each point on the line satisfies these equations. They have the general form \[y = mx + c\]where
  • \(m\) represents the slope of the line, and
  • \(c\) is the y-intercept. This means the point where the line crosses the y-axis.
In this exercise, the line equation is straightforward:\[y = 15\]
This equation signifies a horizontal line cutting through the y-axis at 15. By comparing this line to the parabola, we can find points where they meet (intersect). To do this, we set the y-values of both equations equal. This finds the x-values where the parabola \(y = 2x^2 - 4x - \frac{5}{2}\) matches the line. Substituting these x-values back into the line equation confirms that their y-values are 15, providing the intersection points.

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Most popular questions from this chapter

Show that the ball's path in Example 9 is a parabola by eliminating the parameter in the parametric equations \(x=\left(140 \cos 31^{\circ}\right) t \quad\) and \(\quad y=\left(140 \sin 31^{\circ}\right) t-16 t^{2}\) [Hint: Solve the first equation for \(t\), and substitute the result in the second equation. \(]\)

Find the equation of the ellipse that satisfies the given conditions. Center (7,-4)\(;\) foci on the line \(x=7 ;\) major axis of length \(12 ;\) minor axis of length 5.

Show that the length of the latus rectum of the parabola with equation \(\left.y^{2}=4 p x \text { or } x^{2}=4 p y \text { is } 4|p| . \text { [ Hint: Exercise } 73 .\right]\)

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$x^{2}+y^{2}-4 x-6 y+9=0$$

If \(a, b\) are nonzero constants, show that the graph of \(r=\) \(a \sin \theta+b \cos \theta\) is a circle. \([\text {Hint}: \text { Multiply both sides by } r\) and convert to rectangular coordinates.]
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