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Chapter 11: Problem 11

Graph the parabolas. In each case, specify the focus, the directrix, and the focal width. Also specify the vertex. $$4 x^{2}=7 y$$

Short Answer

Expert verified
Vertex: \((0, 0)\), Focus: \((0, \frac{7}{16})\), Directrix: \(y = -\frac{7}{16}\), Focal width: \(\frac{7}{4}\).

Step by step solution

01

Rearrange the Equation

The given parabola equation is \[4x^2 = 7y\]Rewrite it in the form \[(x-h)^2 = 4p(y-k)\].Divide both sides by 7 to isolate \(y\):\[y = \frac{4}{7}x^2\].This suggests the vertex form centered at the origin \((0, 0)\), with the equation \[(x-0)^2 = \frac{7}{4}(y-0)\].Thus, this is a vertical parabola that opens upward.
02

Identify the Vertex

From the restated equation \[(x-0)^2 = \frac{7}{4}(y-0)\],we can see the vertex \((h, k)\) is at \[(0, 0)\].Thus, the vertex of the parabola is at the origin \((0, 0)\).
03

Determine the Parameter \(p\)

The standard form of the vertical parabola is\[(x-h)^2 = 4p(y-k)\].Comparing with the obtained form \[(x-0)^2=\frac{7}{4}y\],we have \[4p = \frac{7}{4}\],which gives us \[p = \frac{7}{16}\].
04

Find the Focus

The focus of the parabola \[(x-h)^2 = 4p(y-k)\]is located at \((h, k+p)\).Since \(h = 0\), \(k = 0\), and \(p = \frac{7}{16}\),the focus is at \((0, 0 + \frac{7}{16})\) or \((0, \frac{7}{16})\).
05

Find the Directrix

The directrix of the parabola \[(x-h)^2 = 4p(y-k)\]is the line \(y = k - p\).Using \(k = 0\) and \(p = \frac{7}{16}\),the directrix is \[y = 0 - \frac{7}{16} = -\frac{7}{16}\].
06

Determine the Focal Width

The focal width (or latus rectum) of the parabola is given by the absolute value \(|4p|\).Substitute \(p = \frac{7}{16}\) into the equation:\[|4 \times \frac{7}{16}| = \frac{28}{16} = \frac{7}{4}\].This means the focal width is \(\frac{7}{4}\).

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Key Concepts

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

Focus of a Parabola
The focus of a parabola is a key concept in understanding its shape and direction. Located along the axis of symmetry, the focus is a fixed point used to define the parabola. In the equation form \((x-h)^2 = 4p(y-k)\), the focus can be found at \((h, k+p)\). Here,
  • \(h\) is the x-coordinate,
  • \(k\) is the y-coordinate,
  • \(p\) represents the distance from the vertex to the focus.
In our example, \((0, \frac{7}{16})\) is the focus, so the parabola opens upwards centered around this point. The role of the focus makes it crucial for determining the direction in which the parabola opens and helps in constructing the curve accurately.
Directrix of a Parabola
The directrix of a parabola is a straight line that, together with the focus, aids in the geometric definition of the curve. Every point on the parabola is equidistant from the directrix and the focus, which forms the unique shape of the parabola.In the particular form of our problem, \((x-h)^2 = 4p(y-k)\), the directrix is given by the line \(y = k - p\).
  • The constant \(k\) is part of the vertex coordinates,
  • While \(p\) is the distance from the vertex to both the focus and the directrix.
For the equation in question, the directrix is calculated as \(y = -\frac{7}{16}\). This horizontal line below the vertex reflects the parabola鈥檚 opening direction and is crucial for graphing the parabola accurately.
Vertex of a Parabola
The vertex is the pinnacle or nadir of the parabola, depending on its orientation. It is the point at which the parabola changes direction and is located at \((h, k)\) in the vertex form.For our problem, the equation \((x-0)^2 = \frac{7}{4}(y-0)\) indicates that the vertex is at \( (0, 0) \). Here are some pertinent points:
  • The vertex is a reliable starting point for graphing the parabola,
  • It also serves as a reference to calculate both the focus and the directrix.
Since the vertex is at the origin in this case, it simplifies many aspects of the graphing and plotting process.
Latus Rectum
The latus rectum is a segment of the parabola that is important for understanding its width. It is a line segment that runs through the focus, perpendicular to the axis of symmetry, and it helps in visualizing the steepness and spread of the parabola.The length of the latus rectum is given by the absolute value of \(|4p|\).
  • For a vertical parabola, it is calculated using the equation form \((x-h)^2 = 4p(y-k)\).
  • In our example, substituting \(p = \frac{7}{16}\) yields \(|4 \times \frac{7}{16}| = \frac{7}{4}\).
This value indicates how wide the parabola is at its focus and gives an insight into the distribution of the parabola's curvature. Understanding the latus rectum is essential for accurately sketching both narrow and wide parabolic curves.

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Most popular questions from this chapter

Show that the two asymptotes of the hyperbola \(x^{2}-y^{2}=16\) are perpendicular to each other.

Let \(F\) and \(V\) denote the focus and the vertex, respectively, of the parabola \(x^{2}=4 p y .\) If \(\overline{P Q}\) is a focal chord of the parabola, show that $$P F \cdot F Q=V F \cdot P Q$$

In designing an arch, architects and engineers sometimes use a parabolic arch rather than a semicircular arch. (One reason for this is that, in general, the parabolic arch can support more weight at the top than can the semicircular arch.) In the following figure, the blue arch is a semicircle of radius \(1,\) centered at the origin. The red arch is a portion of a parabola. As is indicated in the figure, the two arches have the same base and the same height. Assume that the unit of distance for each axis is the meter. GRAPH CANT COPY (a) Find the equation of the parabola in the figure. (b) Using calculus, it can be shown that the area under this parabolic arch is \(\frac{4}{3} \mathrm{m}^{2} .\) Assuming this fact, show that the area beneath the parabolic arch is approximately \(85 \%\) of the area beneath the semicircular arch. (c) Using calculus, it can be shown that the length of this parabolic arch is \(\sqrt{5}+\frac{1}{2} \ln (2+\sqrt{5})\) meters. Assuming this fact, show that the length of the parabolic arch is approximately \(94 \%\) of the length of the semicircular arch.

A normal to an ellipse is a line drawn perpendicular to the tangent at the point of tangency. Show that the equation of the normal to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) can be written $$a^{2} y_{1} x-b^{2} x_{1} y=\left(a^{2}-b^{2}\right) x_{1} y_{1}$$

This exercise outlines the steps required to show that the equation of the tangent to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) on the ellipse is \(\left(x_{1} x / a^{2}\right)+\left(y_{1} y / b^{2}\right)=1\) (a) Show that the equation \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is equivalent to $$b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$$(1) Conclude that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse if and only if $$b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}$$(2) (b) Subtract equation ( \(2)\) from equation (1) to show that $$b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0$$(3) Equation ( 3 ) is equivalent to equation ( 1 ) provided only that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse. In the following steps, we will find the algebra much simpler if we use equation ( 3 ) to represent the ellipse, rather than the equivalent and perhaps more familiar equation (1). (c) Let the equation of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) be $$y-y_{1}=m\left(x-x_{1}\right)$$ Explain why the following system of equations must have exactly one solution, namely, \(\left(x_{1}, y_{1}\right);\) $$\left\\{\begin{array}{ll}b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0 & (4) \\ y-y_{1}=m\left(x-x_{1}\right) & (5)\end{array}\right.$$ (d) Solve equation ( 5 ) for \(y\), and then substitute for \(y\) in equation (4) to obtain \(b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2} m^{2}\left(x-x_{1}\right)^{2}+2 a^{2} m y_{1}\left(x-x_{1}\right)=0\) (e) Show that equation (6) can be written \(\left(x-x_{1}\right)\left[b^{2}\left(x+x_{1}\right)+a^{2} m^{2}\left(x-x_{1}\right)+2 a^{2} m y_{1}\right]=0\) (f) Equation (7) is a quadratic equation in \(x,\) but as was pointed out earlier, \(x=x_{1}\) must be the only solution. (That is, \(x=x_{1}\) is a double root.) Thus the factor in brackets must equal zero when \(x\) is replaced by \(x_{1}\) Use this observation to show that $$m=-\frac{b^{2} x_{1}}{a^{2} y_{1}}$$ This represents the slope of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) (g) Using this value for \(m,\) show that equation (5) becomes $$b^{2} x_{1} x+a^{2} y_{1} y=b^{2} x_{1}^{2}+a^{2} y_{1}^{2}$$(8) (h) Now use equation (2) to show that equation (8) can be written $$\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$$ which is what we set out to show.
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