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Chapter 11: Problem 20

Find an equation of the following parabolas, assuming the vertex is at the origin. A parabola that opens downward with directrix \(y=6\)

Short Answer

Expert verified
Answer: The equation of the parabola is \(y=\dfrac{-x^2}{18}+\dfrac{3}{2}\).

Step by step solution

01

Understanding parabola definitions

A parabola can be defined as the set of all points that have the same distance to the focus point and the directrix line. Since the vertex is at the origin and the parabola opens downward, this means the focus is above the vertex and the directrix is below it. The vertex will be the midpoint between the focus and the directrix line.
02

Find the focus point and the directrix line equation

Given the directrix line \(y=6\), we can find the focus point by finding the midpoint between the vertex (origin) and the directrix line. The focus point, \(F\), will have the same x-coordinate as the vertex, so \(F = (0, -a)\) for some positive value \(a\). To find the value of \(a\), notice that the distance from the origin to the directrix is 6 units. This means that the focus point is 6 units away from the directrix as well. So, \(a=3\) and the focus point is \(F=(0,-3)\).
03

Set up the equation of the parabola

Let \((x,y)\) be a point on the parabola. According to the parabola definition, the distance from \((x,y)\) to the focus \((0, -3)\) must be equal to the distance from \((x,y)\) to the directrix \(y=6\). We will use the distance formula to set up the equation: Distance from \((x,y)\) to the focus: \(\sqrt{(x-0)^2 + (y-(-3))^2}=\sqrt{x^2+(y+3)^2}\) Distance from \((x,y)\) to the directrix: \(|y-6|\) Now set these two distances equal: \(\sqrt{x^2+(y+3)^2} = |y-6|\)
04

Square both sides to remove the square root

To eliminate the square root, we will square both sides of the equation: \((x^2+(y+3)^2) = (y-6)^2\) Now, we can expand and simplify: \(x^2+y^2+6y+9=y^2-12y+36\)
05

Solve for y

We can eliminate the \(y^2\) terms and rearrange the equation to solve for \(y\): \(x^2+6y+9=-12y+36\) Now, combine the \(y\) terms and move the constant term to the left side: \(x^2+18y-27=0\) Finally, we have the equation of the parabola: \(y=\dfrac{-x^2}{18}+\dfrac{3}{2}\)

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Most popular questions from this chapter

A blood vessel with a circular cross section of constant radius \(R\) carries blood that flows parallel to the axis of the vessel with a velocity of \(v(r)=V\left(1-r^{2} / R^{2}\right),\) where \(V\) is a constant and \(r\) is the distance from the axis of the vessel. a. Where is the velocity a maximum? A minimum? b. Find the average velocity of the blood over a cross section of the vessel. c. Suppose the velocity in the vessel is given by \(v(r)=V\left(1-r^{2} / R^{2}\right)^{1 / p},\) where \(p \geq 1 .\) Graph the velocity profiles for \(p=1,2,\) and 6 on the interval \(0 \leq r \leq R .\) Find the average velocity in the vessel as a function of \(p .\) How does the average velocity behave as \(p \rightarrow \infty ?\)

Let \(R\) be the region bounded by the upper half of the ellipse \(x^{2} / 2+y^{2}=1\) and the parabola \(y=x^{2} / \sqrt{2}\) a. Find the area of \(R\). b. Which is greater, the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or the volume of the solid generated when \(R\) is revolved about the \(y\) -axis?

Find an equation of the following hyperbolas, assuming the center is at the origin. Sketch a graph labeling the vertices, foci, and asymptotes. Use a graphing utility to check your work. A hyperbola with vertices (卤4,0) and foci (卤6,0)

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Sketch the three basic conic sections in standard position with vertices and foci on the \(y\) -axis.
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