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Please help. My textbook does no decent job of explaining this.

 Jun 7, 2019
 #1
avatar+104704 
+1

First one

 

-16x  = y^2      this is a parabola  opening to the left.....we have the form

 

4p ( x - h)  =  (y - k)^2

 

4p ( x - 0)  = (y - 0)^2

 

The vertex is at  (h, k)  = (0,0)

 

We can find "p" [ the distance between the focus and the vertex ] as

 

4p = - 16       divide both sides by 4

p = - 4

 

The focus is given by

 

(0 + p, 0)  =  (0 + (-4), 0)  =  (-4, 0)

 

The directrix is given by

 

x =  - (p) 

 

x= - (-4)

 

x  = 4

 

Here's a graph :  https://www.desmos.com/calculator/cdaauapqz1

 

 

cool cool cool

 Jun 7, 2019
 #2
avatar+104704 
+1

Second one

 

x^2          y^2

___   -     ____   =  1

 25             1

 

We have the form

 

(x - h)^2           (y -k)^2

______   -      ________   =       1         

    a^2                  b^2

 

This is a hyperbola intersecting the x axis  with a center  at (h, k)  = (0, 0)

a = 5      and b  = 1

 

Tve vertices   lie    along the x axis  and  are given by  (a, 0)  and (-a, 0)  = (5, 0) and (-5, 0) 

 

The foci  are given by   ( √[ a^2 + b^2], 0 )    and (- √[a^2 + b^2], 0 )   =

( √ [25 + 1] , 0 )   =  (√26, 0)       and  ( - √26, 0)

 

The asypmtotes are given by :

y = ± (b/a) (x - h) + k     =   

y  = ± (1/5) ( x - 0) + 0    = 

y =  ± (1/5)x 

 

Here's a graph:  https://www.desmos.com/calculator/aabdp29eok

 

cool cool cool

 Jun 7, 2019
 #3
avatar+104704 
+1

Last one

 

x^2         y^2

___  +   ____  =   1   

25          16

 

The " + "  between the terms signals an ellipse

 

Like the hyperbola before, this is centered  at the origin = (0, 0)

 

The major axis is along x  and the minor along y

 

a^2  = 25       b^2  = 16

 

a = 5     b  =  4

 

The vertices here are located at   ( 5, 0) , (-5, 0), (0, 4) and (0, -4)

 

The foci are located on the major axis and  are given by  ( ±√[ a^2 - b^2 ] , 0) =

( ±√[ 25 - 16 ] , 0 )   =   (± √9 , 0 )   =  (3, 0) and (-3, 0)

 

Here's a graph :  https://www.desmos.com/calculator/j579zix3zu

 

cool cool cool

 Jun 7, 2019
 #4
avatar+104704 
0

BTW....here's a resource that I find helpful :

 

https://www.purplemath.com/modules/index.htm

 

 

cool cool cool

 Jun 7, 2019
 #5
avatar+895 
+1

Thank you sooo much CPhill these really helped me. One question though. For the second question, is that formula for asymptotes applicable to all hyperbolas or just this one specifically?

AdamTaurus  Jun 7, 2019
 #6
avatar+104704 
+1

Whenever the hyperbola intersects the x axis  [like the one we did ]

 

The equation of the asymptotes  is

 

y  = ± (b/a) ( x - h) + k       where (h, k) is the center of the hyperbola

 

If the hyperbola intersects the y axis we have the form

 

(y - k)^2        ( x - h)^2

______    -   ________   = 1

 a^2                  b^2

 

And the equation of the asymptotes is

 

y  =  ± (a/b) (x - h) +  k     where (h, k)  is the center

 

 

cool cool  cool

 Jun 7, 2019

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