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Sketch the ellipse, and label the foci, vertices, and ends of the minor axis. (a) \((x+3)^{2}+4(y-5)^{2}=16\) (b) \(\frac{1}{4} x^{2}+\frac{1}{9}(y+2)^{2}-1=0\)

Short Answer

Expert verified
(a) Center (-3, 5), vertices (-7, 5), (1, 5), ends (-3, 3), (-3, 7), foci (-3±2√3, 5). (b) Center (0, -2), vertices (0, 1), (0, -5), ends (2, -2), (-2, -2), foci (0, -2±√5).

Step by step solution

01

Rearrange Equation (a)

First, we rewrite the equation \((x+3)^{2}+4(y-5)^{2}=16\)in standard form for an ellipse. Divide the entire equation by 16 to obtain:\[\frac{(x+3)^{2}}{16} + \frac{4(y-5)^{2}}{16} = 1,\]which simplifies to:\[\frac{(x+3)^{2}}{16} + \frac{(y-5)^{2}}{4} = 1.\]Now, it matches the standard ellipse form \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\).
02

Identify Center, Axes, and Vertices for (a)

The equation \(\frac{(x+3)^{2}}{16} + \frac{(y-5)^{2}}{4} = 1\) reveals that the center \((h, k)\) of the ellipse is at \((-3, 5)\). The values \(a^{2} = 16\) and \(b^{2} = 4\), so \(a = 4\) and \(b = 2\).Since \(a > b\), the major axis is horizontal. Hence, the vertices are at \((-3\pm 4, 5)\), or \((-7, 5)\) and \((1, 5)\). The ends of the minor axis are at \((-3, 5\pm 2)\), or \((-3, 3)\) and \((-3, 7)\).
03

Calculate Foci for (a)

The distance to each focus from the center is \(c\), where \(c^{2} = a^{2} - b^{2}\). Calculating \(c\) for this ellipse:\[c^{2} = 16 - 4 = 12 \]\[c = \sqrt{12} = 2\sqrt{3}.\]The foci are at \((-3\pm 2\sqrt{3}, 5)\).
04

Rearrange Equation (b)

For equation \(\frac{1}{4} x^{2} + \frac{1}{9} (y+2)^{2} - 1 = 0\), rewrite it in the standard form by eliminating the trailing \(-1\):\[\frac{1}{4} x^{2} + \frac{1}{9} (y+2)^{2} = 1.\]Rewrite the equation, multiplying through by 36 to clear fractions:\[9x^{2} + 4(y+2)^{2} = 36,\]which simplifies to:\[\frac{x^{2}}{4} + \frac{(y+2)^{2}}{9} = 1.\]This matches the standard form \(\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\).
05

Identify Center, Axes, and Vertices for (b)

The center of the ellipse \(\frac{x^{2}}{4} + \frac{(y+2)^{2}}{9} = 1\) is at \((0, -2)\). Here, \(a^{2} = 4\) and \(b^{2} = 9\), so \(a = 2\) and \(b = 3\).Since \(b > a\), the major axis is vertical and the minor axis is horizontal. The vertices are \((0, -2\pm 3)\), or \((0, 1)\) and \((0, -5)\). The ends of the minor axis are \((0\pm 2, -2)\), or \((2, -2)\) and \((-2, -2)\).
06

Calculate Foci for (b)

The distance \(c\) to each focus is given by \(c^{2} = b^{2} - a^{2}\). For this ellipse:\[c^{2} = 9 - 4 = 5\]\[c = \sqrt{5}.\]Therefore, the foci are located at \((0, -2\pm \sqrt{5})\).

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Key Concepts

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

Ellipse Sketching
Sketching an ellipse is a wonderful way to visualize its properties and spatial presence. An ellipse resembles a stretched circle, or, as many say, an elongated sphere.
To sketch an ellipse, we rely on its standard equation form, \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), where \( (h, k) \) is the center. Here, \( a \) is the semi-major axis length, and \( b \) is the semi-minor axis.
When sketching, it's vital to identify:
  • The center, \( (h, k) \).
  • The vertices, which are \( (h \pm a, k) \) for a horizontal major axis, and \( (h, k \pm b) \) for a vertical major axis.
  • The co-vertices, lying along the minor axis, at \( (h \pm b, k) \) or \( (h, k \pm a) \).
  • The foci, which lie further from the center along the major axis.
Once you have these key points, draw in the smooth, oval shape of the ellipse, connecting all these points.
Conic Sections
Ellipses are one of the fascinating shapes that stem from the study of conic sections. When we intersect a plane with a cone at an angle that isn't parallel to the base, depending on that angle, the resulting shape could be circular, an ellipse, a parabola, or a hyperbola.
Conic sections, hence, are named for their derivation from cones and are foundational in geometry due to their unique properties.
An ellipse emerges when the intersecting plane cuts through both nappes of a double cone but doesn't pass parallel to the base.
Among conic sections:
  • **Circle**: occurs when the plane is parallel to the cone's base.
  • **Ellipse**: the general form, occurring when intersected at other angles.
  • **Parabola**: emerges when the intersection is parallel to an element of the cone.
  • **Hyperbola**: forms when the plane intersects both conical nappes.
Understanding how these shapes come about highlights their interrelation, deepening one's grasp of geometry.
Each conic section has its specific equation form, showing their distinct characteristics mathematically.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is a fundamental tool in mathematics that allows us to describe geometric shapes using algebra. This branch of mathematics connects algebraic equations with geometric figures in a coordinate plane.
When discussing ellipses within coordinate geometry, each ellipse has a mathematical representation based on a set of coordinates.
Instead of visuals alone, its equation informs us of:
  • The center point of the ellipse.
  • Orientation and length of major and minor axes.
  • Position of the foci, which are integral in defining an ellipse.
The use of coordinate geometry makes it easier to calculate and predict points on an ellipse or distances and areas, with high precision.
It allows for a deeper exploration of properties such as eccentricity, symmetry, and even rotational dynamics of these shapes inside a coordinate system. This analytical approach helps illuminate how ellipses and other conic sections fit into broader mathematical landscapes, expanding both theoretical and practical understanding.

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Most popular questions from this chapter

Find the coordinates of all points on the hyperbola $$4 x^{2}-y^{2}=4$$ where the two lines that pass through the point and the foci are perpendicular.

Discuss how a hyperbola's eccentricity \(e\) affects the shape of the hyperbola. How is the shape affected as \(e\) approaches \(1 ?\) As \(e\) approaches \(+\infty\) ? Draw some pictures to illustrate your conclusions.

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. Find the area of the surface generated by revolving \(x=t^{2}\), \(y=3 t(0 \leq t \leq 2)\) about the \(x\) -axis.

Show that the graph of the given equation is a hyperbola. Find its foci, vertices, and asymptotes. $$ 17 x^{2}-312 x y+108 y^{2}-900=0 $$

Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter. $$ x=\theta+\cos \theta, \quad y=1+\sin \theta ; \theta=\pi / 6 $$

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