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**Chapter 10 Section 5 Hyperbola**

Algebra II Chapter 10 Section 5 Hyperbola

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Hyperbola The set of all points that have the difference of the distance to 2 focus points is a given constant

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Parts of a hyperbola Asymptotes Foci Transverse Axis Vertices

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**Standard Form of an Hyperbola**

If a & b > 0 the Transverse axis is Horizontal If a & b < 0 the Transverse axis is Vertical

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Put into standard Form

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Put into standard Form

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In standard form The +-value for a (square root of the denominator of x2) are the vertex’s x if the Transversal Axis is horizontal, values (-a,0) and (a,0) The +-value for b (square root of the denominator of y2) are the vertex’s y if the Transversal Axis is Vertical, values (0,-b) and (0,b)

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To find the Asymptote

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**Find the Asymptotes and Vertices**

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**(4,0) and (-4,0) a = 4 b = 5 Horizontal Transverse Axis**

So Vertices are at +- (a,0) (4,0) and (-4,0) Asymptotes are at

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**Given a vertex at (-3, 0) and an asymptote of y = 4x**

Find the equation of hyperbola

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**Foci Foci are always on the transverse axis**

Foci are equidistant from the minor axis The distances to a focus from the origin is equal to the distance along the asymptote, with a coordinate = to the vertex’s non zero coordinate

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**Find the focus of an Hyperbola with its center on the origin**

If c is the distance from the center to the focus Find c using c2 = a2 + b2 if your transverse axis is Horizontal your foci are at (+-c ,0) if your transverse axis is Vertical your foci are at (0 ,+-c)

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Find the foci of a & b are + so we have a horizontal Transverse axis (y = 0) c2 = a2 + b2 c2 = c2 = 25 c = 5 So the coordinates of the foci are at (5,0) and (-5,0)

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Find the Foci

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**Find the equation given a vertex and a focus**

Find the other vertex using c2 = a2 + b2 Substitute in the 2 values you know and solve for the 3rd Remember the foci will lie along the Transverse axis c is the distance of the focus to the origin Once you have the 2 vertices, take the square roots to get a and b & substitute them in

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**Given a focus of (10,0) and a vertex at (-6,0) find the equation of the Hyperbola**

The transverse axis is Horizontal since the vertex and focus lie on the x axis Since the vertex is at (0,-6) a2 = 36 Since the focus is at (0,10) c2 = 100 Substitute a2 and c2 into a2 + b2 = c2 and solve for b 36 + b2 = 100 b2 = 64 b= 8 Substitute a2 & b2 into To get

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**Given a focus of (13,0) and a vertex at (5,0) find the equation of the ellipse**

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Do Now! Page 566 Problems 2 – 18 even, find the vertex, foci and asymptotes as well as sketch a graph 23 – 26 all 30, 31, 36, 40, 42,

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Section 9.2 The Hyperbola. Overview In Section 9.1 we discussed the ellipse, one of four conic sections. Now we continue onto the hyperbola, which in.

Section 9.2 The Hyperbola. Overview In Section 9.1 we discussed the ellipse, one of four conic sections. Now we continue onto the hyperbola, which in.

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