⧼exchistory⧽
16 exercise(s) shown, 0 hidden
BBy Bot
Nov 03'24
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[/math]
Simplify [math]\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} = 2a[/math] and show that it results in the equation [math]\frac{x^2}{z^2} - \frac{y^2}{c^2-a^2} = 1[/math].
BBy Bot
Nov 03'24
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[/math]
Describe and sketch the graph of each of the following equations. Label the foci and vertices and sketch the asymptotes.
- [math]\frac{x^2}{9} - \frac{y^2}{16} = 1[/math]
- [math]\frac{x^2}{18} - \frac{y^2}{7} = 1[/math]
- [math]\frac{x^2}{25} - \frac{y^2}{144} = 1[/math]
- [math]\frac{x^2}{225} - \frac{y^2}{64} = 1[/math]
- [math]25y^2 - 9x^2 = 225[/math]
- [math]xy = 8[/math]
- [math]xy = -6[/math].
BBy Bot
Nov 03'24
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[/math]
It has been shown that the distance between a point on the hyperbola [math]\frac{x^2}{a^2} - \frac{y^2}{c^2-a^2} = 1[/math] and the focus [math](c,0)[/math] is [math]\left| \frac ca x - a \right|[/math]. Call this distance [math]d_1[/math].
- Show that the distance [math]d_2[/math] between a point on the hyperbola and the focus [math](-c,0)[/math] is [math]\left| \frac ca x + a \right|[/math].
- Show that [math]x \geq a[/math] for a point on the right branch of the hyperbola and that for such a point [math]d_1 = \frac ca x - a[/math] and [math]d_2 = \frac ca x + a[/math].
- Show that [math]x \leq -a[/math] for a point on the left branch of the hyperbola and that for such a point [math]d_1 = -\frac ca x + a[/math] and [math]d_2 = -\frac ca x - a[/math].
- Hence show that the graph of [math]\frac{x^2}{a^2} - \frac{y^2}{c^2-a^2} = 1[/math] contains only those points which satisfy the locus definition of hyperbola.
BBy Bot
Nov 03'24
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[/math]
Write an equation for the hyperbola with horizontal and vertical axes satisfying the given conditions.
- Foci at [math](-5,0)[/math] and [math](5,0)[/math]. Transverse axis of length [math]6[/math].
- Center at the origin. Transverse axis horizontal and of length [math]4[/math], conjugate axis vertical and of length [math]12[/math].
- Center at the origin. Transverse axis vertical and of length [math]6[/math]. Passing through [math](1, 2\sqrt3)[/math].
- Center at the origin. Passing through [math](1,5)[/math] and [math](2,7)[/math].
BBy Bot
Nov 03'24
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[/math]
- lab{3.4.5a} Show that the graph of [math](x-3)(y+2) = 10[/math] is an equilateral hyperbola with center at [math](3,-2)[/math] and asymptotes [math]x=3[/math] and [math]y=-2[/math].
- Sketch the graph described in \ref{ex3.4.5a}.
BBy Bot
Nov 03'24
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[/math]
Describe and sketch the graph of each of the following equations. Label the foci and vertices and sketch the asymptotes.
- [math]\frac{(x-1)^2}{16} - \frac{(y+3)^2}{9} = 1[/math]
- [math]\frac{(y+3)^2}{9} - \frac{(x-1)^2}{16} = 1[/math]
- [math](x-1)(y+3) = -12[/math]
- [math]144(x+4)^2 - 25(y+2)^2 = 3600[/math]
- [math]2xy - 4x - 4y - 17 = 0[/math]
- [math]x^2 - y^2 - 2x + 2y - 2 = 0[/math]
- [math]y^2 - 4x^2 + 4y - 16x - 28 = 0[/math].
BBy Bot
Nov 03'24
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[/math]
Hyperbolas with the same pair of foci are said to be confocal. Show that the following equations have confocal hyperbolas for their graphs and sketch them all on the same set of axes.
- [math]x^2 - \frac{y^2}{24} = 1[/math]
- [math]\frac{x^2}{9} - \frac{y^2}{16} = 1[/math]
- [math]\frac{x^2}{16} - \frac{y^2}{9} = 1[/math]
- [math]\frac{x^2}{24} - y^2 = 1[/math].
BBy Bot
Nov 03'24
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[/math]
The line segment which passes through a focus, is perpendicular to the transverse axis extended, and has its endpoints on the hyperbola is called a latus rectum.
- Find the length of a latus rectum of the hyperbola [math]9x^2 - 16y^2 = 144[/math].
- Find the length of a latus rectum of the hyperbola [math]b^2x^2 - a^2y^2 = a^2b^2[/math].
- Show that both latera recta of a hyperbola are the same length.
BBy Bot
Nov 03'24
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[/math]
Write equations of the directrices and find the eccentricity of each of the following hyperbolas.
- [math]9x^2 - 16y^2 = 144[/math]
- [math]144y^2 - 25x^2 = 3600[/math].
BBy Bot
Nov 03'24
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[/math]
Assume that [math]0 \lt a \lt c[/math].
- lab{3.4.10a} Find the distance between [math](x,y)[/math] and [math](-c,0)[/math].
- lab{3.4.10b} Find the distance between [math](x,y)[/math] and the line [math]x = -\frac{a^2}c[/math].
- Find the locus of points [math](x,y)[/math] such that the ratio between the distance in \ref{ex3.4.10a} and the distance in \ref{ex3.4.10b} is a constant [math]\frac ca[/math].