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A circle has the same center as an ellipse and passes through the foci \(F_1\) and \(F_2\) of the ellipse. The two curves intersect in 4 points. Let P be any point of intersection. If the major axis of the ellipse is 15 and the area of triangle \(PF_1 F_2\) is 26, compute the distance between the foci.

 Jul 31, 2019
 #1
avatar+104937 
+2

Let the center  =  (0, 0)

Let the major axis  be along the x axis

a =  7.5

c^2 = a^2 - b^2

c^2 = 7.5^2 - b^2   ⇒  b^2  = 7.5^2 - c^2

2c  =  the base of  triangle  PF1F2

Radius of circle  = c

Area of triangle  =   (1/2) (2c) ( height)  = 26     ⇒ c (height) = 26  ⇒  height = 26/c 

So.....P is on the ellipse and circle

Let P  =  ( x, 26/c)

And the equation of the circle = x^2 + y^2  = c^2   ⇒  x^2 + (26/c)^2  = c^2  ⇒  x^2  = c^2 - (26/c)^2

 

Using the equation for an ellipse :

x^2/ a^2  + y^2 /b^2  = 1       and substituting, we have :

 

[c^2 - (26/c)^2] / 7.5^2  +  (26/c)^2 / [ 7.5^2 - c^2]   =  1

 

Solving this for  c is a little difficult  but we get that c ≈  5.5

And the height of the triangle  = 26/c  =  26/5.5  = 52/11    =  y coordinate of P

And the x ooordinate for P  = sqrt (5.5^2 - (26/5.5)^2 ) ≈  2.81

 

So  P  = (2.81, 52/11)

 

So...the distance between the foci  2c = 2(5.5) ≈  11

 

Here's a pic :

 

 

cool cool cool

 Jul 31, 2019
 #2
avatar+105689 
+1

Thanks Chris,

One of these days I am going to study conics properly.  I don't know much about them.   sad

Melody  Jul 31, 2019
 #3
avatar+104937 
+1

Well......if it makes you feel any better.....I have to review some of them, too.....particularly the hyperbola  !!!

 

 

cool cool cool

CPhill  Jul 31, 2019

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