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Determine the foci, the eccentricity, and the directrices for each ellipse and hyperbola. (a) \(25 x^{2}+36 y^{2}=900\) (b) \(25 x^{2}-36 y^{2}=900\)

Short Answer

Expert verified
(a) Foci: \((\pm\sqrt{11}, 0)\), Eccentricity: \(\frac{\sqrt{11}}{6}\), Directrices: \(x = \pm\frac{36\sqrt{11}}{11}\). (b) Foci: \((\pm\sqrt{61}, 0)\), Eccentricity: \(\frac{\sqrt{61}}{6}\), Directrices: \(x = \pm\frac{36\sqrt{61}}{61}\).

Step by step solution

01

Rewrite the Ellipse Equation as a Standard Form

Given the equation \(25x^2 + 36y^2 = 900\) for the ellipse.Divide every term by 900 to get:\[ \frac{x^2}{36} + \frac{y^2}{25} = 1 \]This matches the standard form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a^2 = 36\) and \(b^2 = 25\). Therefore, \(a = 6\) and \(b = 5\).
02

Find the Foci and Eccentricity of the Ellipse

Foci for an ellipse are given by \((\pm c, 0)\), and the relationship between \(a\), \(b\), and \(c\) (the distance from the center to each focus) is \(c = \sqrt{a^2-b^2}\).Calculate \(c\):\[ c = \sqrt{36 - 25} = \sqrt{11} \]So, the foci are \((\pm \sqrt{11}, 0)\).Eccentricity \(e\) is computed as \(e = \frac{c}{a}\):\[ e = \frac{\sqrt{11}}{6} \]
03

Find the Directrices of the Ellipse

The directrices for the ellipse are given by \(x = \pm \frac{a^2}{c}\).Calculate \(\frac{a^2}{c}\):\[ \frac{36}{\sqrt{11}} = \frac{36\sqrt{11}}{11} \] (rationalizing the denominator)Thus, the directrices are \(x = \pm \frac{36\sqrt{11}}{11}\).
04

Rewrite the Hyperbola Equation as a Standard Form

Given the equation \(25x^2 - 36y^2 = 900\) for the hyperbola.Divide every term by 900 to get:\[ \frac{x^2}{36} - \frac{y^2}{25} = 1 \]This matches the standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where \(a^2 = 36\) and \(b^2 = 25\). Therefore, \(a = 6\), and \(b = 5\).
05

Find the Foci and Eccentricity of the Hyperbola

Foci for a hyperbola are given by \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\).Calculate \(c\):\[ c = \sqrt{36 + 25} = \sqrt{61} \]So, the foci are \((\pm \sqrt{61}, 0)\).Eccentricity \(e\) is computed as \(e = \frac{c}{a}\):\[ e = \frac{\sqrt{61}}{6} \]
06

Find the Directrices of the Hyperbola

The directrices for the hyperbola are given by \(x = \pm \frac{a^2}{c}\).Calculate \(\frac{a^2}{c}\):\[ \frac{36}{\sqrt{61}} = \frac{36\sqrt{61}}{61} \]Thus, the directrices are \(x = \pm \frac{36\sqrt{61}}{61}\).

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Key Concepts

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

Ellipse
An ellipse is a type of conic section that looks like a stretched circle. You get an ellipse when you slice a cone at an angle that isn't quite perpendicular to the base, but not steep enough to intersect the opposite cone. It can be described mathematically by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(a\) and \(b\) are the distances from the center to the ellipse's edges along the x and y axes, respectively.

The essence of an ellipse is that any point on its outline is the same total distance from two specific points inside the shape. These points are called foci (singular: focus). An ellipse has many interesting properties and uses: from the shape of planetary orbits to the design of reflective surfaces in headlights and satellite dishes. Because of its unique symmetry and form, the study of ellipses is fundamental in fields ranging from engineering to astronomy.
Hyperbola
A hyperbola is another type of conic section, but its shape is formed by slicing through a cone at a sharper angle. This split creates two mirrored, arch-like curves. The standard equation for a hyperbola in orientation with x-axis opening is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \(a\) and \(b\) define the distances related to the axes, similar to an ellipse, but their arrangement signifies that the curve extends infinitely.

Unlike an ellipse, which has a closed shape, a hyperbola opens up infinitely outward. In real-world situations, hyperbolas are found in structures like cooling towers and can also describe the paths of objects in gravitational fields if they are moving fast enough to escape the gravitational pull. This intriguing property makes hyperbolas essential in physics, especially in the study of celestial mechanics and optics.
Foci
The foci of both ellipses and hyperbolas are fundamental. In the case of an ellipse, the foci lie on the major axis, and the sum of the distances from any point on the ellipse to the two foci is constant. This means if you took a piece of string, pinned it at the two foci, and pulled it taut with a pencil, you could draw an ellipse as you move around.

For hyperbolas, the concept is slightly different: the difference in distances from any point on the hyperbola to the two foci remains constant. This property dictates the curve's shape, maintaining symmetry. You can calculate the distance \(c\) from the center to the foci using the equations for \(c\), based on whether you are dealing with an ellipse \(c = \sqrt{a^2-b^2}\) or a hyperbola \(c = \sqrt{a^2+b^2}\). These unique properties help identify the type and properties of the conic section.
Eccentricity
Eccentricity is a measure of how much a conic section deviates from being circular. For an ellipse, the eccentricity \(e\) is given by \(e = \frac{c}{a}\). Here, \(c\) is the distance from the center to a focus, and \(a\) is the distance from the center to a vertex on the major axis. The eccentricity of an ellipse ranges from 0 (a perfect circle) to less than 1, as the shape becomes more elongated.

In hyperbolas, eccentricity is always greater than 1, reflecting their open-ended curves, calculated in a similar way: \(e = \frac{c}{a}\). The greater the eccentricity, the "flatter" or more "stretched out" the shape is. This information is crucial in various scientific applications, such as determining the orbits of planets, which can be slightly elliptical, or assessing the paths of comets with hyperbolic trajectories.
Directrices
Directrices are lines associated with conic sections that serve as reference points just like foci. For ellipses, the directrices help in defining the shape, placed perpendicular to the major axis. Calculating the position of the directrices involves using the formula \(x = \pm \frac{a^2}{c}\). They are primarily utilized in the context of ellipses and hyperbolas to better understand and define these curves.

For hyperbolas, the directrices have the same formula as ellipses, and similarly play a role in determining the curve's properties. Though often less famous than foci, the use of directrices in conjunction with eccentricity and the property of distances is crucial in solving many mathematical problems involving conic sections, enhancing our understanding of their geometric nature.

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Most popular questions from this chapter

Find the equation of the parabola satisfying the given conditions. In each case, assume that the vertex is at the origin. The focus is (0,3)

By solving the system $$\left\\{\begin{array}{l}y=3 x-1 \\\16 x^{2}-9 y^{2}=144\end{array}\right.$$ show that the line and the hyperbola intersect in exactly one point. Draw a sketch of the situation. This demonstrates that a line that intersects a hyperbola in exactly one point does not have to be a tangent line.

In this exercise we graph a hyperbola in which the axes of the curve are not parallel to the coordinate axes. The equation is \(x^{2}+4 x y-2 y^{2}=6\) (a) Use the quadratic formula to solve the equation for \(y\) in terms of \(x .\) Show that the result can be written \(y=x \pm \frac{1}{2} \sqrt{6 x^{2}-12}\) (b) Graph the two equations obtained in part (a). Use the standard viewing rectangle. (c) It can be shown that the equations of the asymptotes are \(y=(1 \pm 0.5 \sqrt{6}) x .\) Add the graphs of these asymptotes to the picture that you obtained in part (b). (d) Change the viewing rectangle so that both \(x\) and \(y\) extend from -50 to \(50 .\) What do you observe?

(a) Verify that the point \(P(6,4 \sqrt{3})\) lies on the hyperbola \(16 x^{2}-9 y^{2}=144\) (b) In Example \(1,\) we found that the foci of this hyperbola were \(F_{1}(-5,0)\) and \(F_{2}(5,0) .\) Compute the lengths \(F_{1} P\) and \(F_{2} P\), where \(P\) is the point \((6,4 \sqrt{3})\) (c) Verify that \(\left|F_{1} P-F_{2} P\right|=2 a\)

Let \(P\) denote the point (8,8) on the parabola \(x^{2}=8 y,\) and let \(\overline{P Q}\) be a focal chord. (a) Find the equation of the line through the point (8,8) and the focus. (b) Find the coordinates of \(Q\) (c) Find the length of \(\overline{P Q}\). (d) Find the equation of the circle with this focal chord as a diameter. (e) Show that the circle determined in part (d) intersects the directrix of the parabola in only one point. Conclude from this that the directrix is tangent to the circle. Draw a sketch of the situation.

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