Suppose points F1, F2 are the left and right foci of the ellipse x216+y24=1 respectively, and point P is on line l:x−3y+8+23=0 . When ∠F1PF2 reaches the maximum, then the value of ratio |PF1||PF2| is

∠F1PF2 reaches the maximum only if the circle through points F1, F2, P is tangent to the line l at P. Now suppose l intercepts the x-axis at point A(−8−23,0) . Then , ∠APF1=∠AF2P and that means ΔAPF1~ΔAF2P . So |PF1||PF2|=|AP||AF2| .By using the power of point’s theorem, we have |AP|2=|AF1|.|AF2|. As F1(−23,0),F2(23,0),A(−8−23,0), So |AF1|=8,|AF2|=8+43 Then we get |PF1||PF2|=|AF1||AF2|=88+43 =4−23=3−1 .

Suppose points F1, F2 are the left and right foci of the ellipse x216+y24=1 respectively, and point P is on line l:x−3y+8+23=0 . When ∠F1PF2 reaches the maximum, then the value of ratio |PF1||PF2| is

∠F1PF2 reaches the maximum only if the circle through points F1, F2, P is tangent to the line l at P. Now suppose l intercepts the x-axis at point A(−8−23,0) . Then , ∠APF1=∠AF2P and that means ΔAPF1~ΔAF2P . So |PF1||PF2|=|AP||AF2| .By using the power of point’s theorem, we have |AP|2=|AF1|.|AF2|. As F1(−23,0),F2(23,0),A(−8−23,0), So |AF1|=8,|AF2|=8+43 Then we get |PF1||PF2|=|AF1||AF2|=88+43 =4−23=3−1 .

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Suppose points F1, F2 are the left and right foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$ respectively, and point P is on line $l : x - \sqrt{3} y + 8 + 2 \sqrt{3} = 0$ . When $∠ F_{1} P F_{2}$ reaches the maximum, then the value of ratio $\frac{| P F_{1} |}{| P F_{2} |}$ is

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$\sqrt{3} - 1$

$\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

$\sqrt{3} + 1$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

answer is B.

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Detailed Solution

$∠ F_{1} P F_{2}$ reaches the maximum only if the circle through points F1, F2, P is tangent to the line $l$ at P. Now suppose $l$ intercepts the x-axis at point $A (- 8 - 2 \sqrt{3}, 0)$ . Then , $∠ A P F_{1} = ∠ A F_{2} P$ and that means $Δ A P F_{1} ~ Δ A F_{2} P$ . So $\frac{| P F_{1} |}{| P F_{2} |} = \frac{| A P |}{| A F_{2} |}$
.
By using the power of point’s theorem, we have ${| A P |}^{2} = | A F_{1} | . | A F_{2} | .$

As $F_{1} (- 2 \sqrt{3}, 0), F_{2} (2 \sqrt{3}, 0), A (- 8 - 2 \sqrt{3}, 0),$ So $| A F_{1} | = 8, | A F_{2} | = 8 + 4 \sqrt{3}$

Then we get $\frac{| P F_{1} |}{| P F_{2} |} = \sqrt{\frac{| A F_{1} |}{| A F_{2} |}} = \sqrt{\frac{8}{8 + 4 \sqrt{3}}}$ $= \sqrt{4 - 2 \sqrt{3}} = \sqrt{3} - 1$
.