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An equilateral triangle ABC is inscribed in the ellipse x^2/a^2+y^2/b^2=1, so that B is at (0,b), and line AC is parallel to the x-axis, as shown below. Also, foci F_1 and F_2 lie on sides line BC and line AB, respectively. Determine the value of AB/(F_1*F_2).

Aug 28, 2020

#1
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The answer works out to 13/8.

Aug 28, 2020
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I would appreciate if you gave a solution following your statement. Thanks in advance.

Aug 28, 2020
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what do you mean by  F_1*F_2

they are points, not distances so it doesn't make sense to me.......

Aug 29, 2020
#4
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I apologize for the misunderstanding, as I meant the length between the two points F_2 and F_1.

DragonLord  Aug 29, 2020
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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 AC

triangle BXF2    and  triangle BMA   are similar tiranges and they are both 30, 60 90 degree triangles.

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

I want the focal length (c) to be 1 and

I want B to be   $$(0,\sqrt3)$$

$$c^2=a^2-b^2\\ 1=a^2-3\\ 4=a^2\\ a=2$$

So the ellipse that I will use is

$$\frac{x^2}{4}+\frac{y^2}{3}=1\\ \frac{y^2}{3}=1-\frac{x^2}{4}$$

Equation of line AB

$$y=\sqrt3\;x+\sqrt3\\ y=\sqrt3(x+1)\\ y^2=3(x+1)^2\\ \frac{y^2}{3}=(x+1)^2\\$$

Solve simultaneously

$$1-\frac{x^2}{4}=(x+1)^2\\ -\frac{x^2}{4}=x^2+2x\\ -x^2=4x^2+8x\\ 5x^2+8x=0\\ x(5x+8)=0\\ x=0\quad or \quad x=-\frac{8}{5}$$

So the x value for A is   -8/3

$$AC=AB=BC=\frac{16}{5}\\ F_2F_1=2\\ \frac{AB}{F_1F_2}=\frac{16}{10}=\frac{8}{5}$$

Aug 30, 2020
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Ohh, I see! Thank you Melody!

Sep 5, 2020