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:

(x-c)²+y²=p².

When the circle intersects the ellipse, we have:

(x-c)²+(a²-c²)(1-x²/a²)=p²,

x²-2cx+c²+a²-x²-c²+c²x²/a²=p²,

-2cx+a²+c²x²/a²=p²,

c²x²-2a²cx+a⁴-a²p²=0.

x=(2a²c±√(4a⁴c²-4c²(a⁴-a²p²))/(2c²),

x=(2a²c±√(4a²c²p²))/(2c²),

x=(2a²c±2acp)/2c²=(a²±ap)/c.

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):

((a²±3a)/c-c)/3=(c-(a²±5a)/c)/5,

(5a²±15a)/c-5c=(3c-3a²∓15a)/c,

(8a²+15a(±1±1))/c-8c=0,

(8a²+15a(±1±1))-8c²=0,

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:

((a²±4a)/c-c)/4=(c-(a²±pa)/c)/p.

(pa²±4pa)/c-pc=4c-(4a²±4pa)/c,

((p+4)a²+4pa(±1±1)-(p+4)c²=0,

(p+4)b²+4pa(-2)=0,

b²p+4b²=8pa,

4b²=p(8a-b²)

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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