Suppose A, B, C and D are points on an ellipse such that segments AB and CD intersect at a focus F. Given that AF = 3, CF = 4 and BF = 5, what is DF? (Note: Use of trigonometric ratios are restricted)

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Equation of ellipse: x²/a²+y²/b²=1.

If F(c,0) is a focus, where c²=a²-b², then

x²/a²+y²/(a²-c²)=1, y²=(a²-c²)(1-x²/a²).

Circle centre F, radius=p:


When the circle intersects the ellipse, we have:








To find the x coordinate of A, circle C₁ radius p=AF=3, x=(a²±3a)/c.

Similarly the x coordinate of B, circle C₂ radius p=BF=5, x=(a²±5a)/c.

The cosine of the inclination of AF (and therefore AB) is (x-c)/3=((a²±3a)/c-c)/3, when x>c, and of BF is (c-(a²±5a)/c)/5 (x<c). When these two cosines are equal  (AB is a chord):





8(a²-c²)+15a(±1±1)=8b²+15a(±1±1)=0, because a²-c²=b² by definition.

So 8b²=-15a(±1±1). The only feasible solution is 8b²=-15a(-2)=30a, b²=15a/4.

So the equation of the ellipse is:

x²/a²+4y²/15a=1. Also, c²=a²-15a/4=a(a-3.75).

That means a>3.75. (The circle must be large enough to intersect the ellipse.)

The cosine of the inclination of CF is ((a²±4a)/c-c)/4 and of DF (unknown)  is (c-a²±pa)/c)/p, because the x coordinates of C and D must be on either side of the focus.

Since these must be equal:







p=4b²/(8a-b²)=15a/(8a-3.75a)=15/4.25=60/17, substituting b²=15a/4.

Therefore DF=60/17=3.5294 approx. does not depend on the shape of the ellipse.



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