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# Ellipses

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An equilateral triangle $PQR$ is inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ so that $Q$ is at $(0,b),$ and $\overline{PR}$ is parallel to the $x$-axis, as shown below. Also, foci $F_1$ and $F_2$ lie on sides $\overline{QR}$ and $\overline{PQ},$ respectively. Find $\frac{PQ}{F_1 F_2}.$ Dec 29, 2020

### 2+0 Answers

#1
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Since it is a ratio question you can use a convenient ellipse.

Let X be the centre and M be the midpoint of PR

triangle QXF2    and  triangle QMP   are similar tiranges and they are both 30, 60 90 degree triangles.

So their sides are in the ration 1:2:sqrt3

We can let the focal length (c) to be 1 and we can Q be (0,1).  Then a = 2.

From here, with routine calculations, you can work out that PQ/F_1 F_2 = 13/8.

Dec 29, 2020
#2
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For convenience let  Q   = (0, 1)

Let the center lie at (0,0)

Note that in the equation of the ellipse, "b"    = 1

Then  1/2  the  focal length  =  1/sqrt (3)

So F1F2  = 2/sqrt (3) =   (2/3)sqrt (3)

We can find "a"   as  sqrt  [ 1^2 + (1/sqrt(3))^2  ]  = sqrt (1 + 1/3)   =  sqrt [ 4/3 ] =  2/sqrt (3)

a^2 =  4/3

A line from Q through R  will have a slope of  tan (-60)  =  -sqrt (3)

So.....the equation of this line is   y  = -sqrt (3)  +  1      (1)

To find the positive x values  of R, sub  (1)  into the equation of the ellipse

x^2 / ( 4/3)  +   [ -sqrt (3)x  + 1]^2  =  1

(3/4)x^2  +  3x^2  -2sqrt (3)x  + 1   = 1

(15/4)x^2  -  2sqrt (3)  x  =  0

x   ( 15/4x -  2sqrt (3)  )  =  0

The second  factor set to 0  will give ys the  x coordinate of R

(15/4)x  - 2sqrt (3)   =  0

x = (4/15)2sqrt (3)

x =  (8/15)sqrt (3)

And the y coordinate  of  R = -sqrt (3) (8/15)sqrt (3)  + 1  =  -24/15 + 1  =  -9/15   = -3/5

And  QR  = PQ =   sqrt   [ (8/15sqrt (3)^2   +  (1 - -3/5)^2  ]    =

sqrt[ (8/5)^2  ( 1/3 + 1 ]  =  (8/5)sqrt [ 1 +1/3  ]  =  (8/5) sqrt  ( 4/3)  =  (16/5) sqrt (1/3)  = (16sqrt (3) / 15)  =

(16/15)sqrt (3)

So

PQ   / F1F2  =     (16/15) sqrt (3)    / [ (2/3)sqrt (3)  ]  =    (16/15) (3/2)  =  8/5   Dec 30, 2020