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avatar+91 

The graph of \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]has its foci at $(0,\pm 4),$ while the graph of \[\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\]has its foci at $(\pm 6,0).$ Compute the value of $|ab|.$

 Jul 10, 2019
 #1
avatar+111321 
+2

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]has its foci at $(0,\pm 4) \)

 

\(\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\]has its foci at $(\pm 6,0).\)

 

Find  l ab l

 

The first equation is an ellipse  with its center at (0, 0) and its major axis along y 

The equation for the focus  is    (0 , 0 ± c)  where  c = ±4

And we have that   a^2 - b^2  = c^2  ⇒  a^2 - b^2  = 16   (1)

 

The second equation is a hyperbola with its center at (0, 0)  and its major axis along x

The equation for the focus is  (0 ± c, 0)  where c = ±6

And we  have that a^2 + b^2  = c^2  ⇒  a^2 + b^2 =  36   (2)

 

Add  (1) and (2)   and we have that

 

2a^2  =  52

a^2  =  26   ⇒    a =  ±√26

 

And  using (2)

 

26 + b^2  = 36

b^2  =  10 ⇒  b = ±√10  

 

So

l ab l  =  l √26 * √10 l  =   l √ 2 * √13 * √2 * √5  l  =    2√65

 

 

cool cool cool

 

 

cool cool cool

 Jul 10, 2019
 #2
avatar+1694 
+6

Sorry, CPhill, just wondering if you knew how to space in latex.

 

If you do not know, you would put a \ and a space at the end of each word in the latex section, hence,

has\ its\ foci\ at

which would equal to:

\(has\ its\ foci\ at\)

 

\(tommarvoloriddle \)

tommarvoloriddle  Jul 11, 2019
 #3
avatar+111321 
0

THX, tom  for that helpful hint !!!

 

 

cool cool cool

CPhill  Jul 11, 2019
 #4
avatar+1694 
+5

np, cphill!

 

: )

tommarvoloriddle  Jul 11, 2019

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