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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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Hyperbolas.

Hyperbolas.

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