Set \ (F_1, F_2 \) is elliptical \ (\ dfrac {x ^ 2 } {a ^ 2} + \ dfrac {y ^ 2} {b ^ 2} = 1 (a> b> 0) \) about focus point \ (P \) on the ellipse, and \ (\ angle F_1PF_2 = \ dfrac {\ PI. 3} {} \) , \ (\ Triangle F_1PF_2 \) of the circumscribed circle of the radius of the inscribed circle to its radius ratio is \ (2: 1 \) .
\ ((1) \) required eccentricity of the ellipse \ (E \) ;
\ ((2) \) disposed \ (AB \) is an ellipse perpendicular to the \ (X \) axis string, \ (C \) coordinates of \ ((3,0) \) , the straight line \ (the BC \) ellipse at point \ (E \) , when the straight line \ (the AE \) constant through the point \ ( \ left (\ dfrac. 4 {} {}. 3, 0 \ right) \) , find the equation of the ellipse.
Analysis:
\ ((1) \) disposed \ (\ triangle F_1PF_2 \) of the circumscribed circle and the inscribed circle radii \ (R & lt, R & lt \) , provided \ (C \) is the focal length of a half ellipse, sinusoidal by theorem, \ [R = \ dfrac {1
} {2} \ cdot \ dfrac {| F_1F_2 |}. {\ sin \ angle F_1PF_2} = \ dfrac {2 \ sqrt {3} c} {3} \] an elliptical focus triangle area formula indicates, \ (\ triangle F_1PF_2 \) area \ [S _ {\ triangle F_1PF_2 } = b ^ 2 \ tan \ dfrac {\ angle F_1PF_2} {2} = \ dfrac {\ sqrt {3}} { 3} b ^ 2 = \ dfrac
{\ sqrt {3}} {3} (a ^ 2-c ^ 2). \] is easy to know \ (\ triangle F_1PF_2 \) of the perimeter \ [L _ {\ triangle F_1PF_2 } = | PF_1 | + | PF_2
| + | F_1F_2 |. = 2 (a + c) \] Calcd from the other area method \ (\ triangle F_1PF_2 \) of the inscribed circle radius of length \ [r = \ dfrac { 2S _ {\ triangle F_1PF_2}} {L _ {\ triangle F_1PF_2}} = \ dfrac {\ sqrt {3}} {3} \ left (ac \ right). \] and the \ (R: r = 2: 1 \) can be obtained \ [E = \ dfrac {C} = {A} \ dfrac 12. The \]
\ ((2) \) method a as shown, is connected \ (the AC \) Since \ (A, B \) on \ (X \) axis of symmetry, the linearly \ (the BC \) and \ (the AC \) also on the straight line \ (X \) axis of
He said to \ (the AC \) another intersection ellipse \ (M \) also \ (E \) on (X \) \ axisymmetric, so \ (M, G, B \ ) are collinear. wherein \ (G \) coordinate is \ [G \ left (\ dfrac
{4} {3}, 0 \ right), \] binding \ ((1) \) may be provided elliptic equation \ (\ dfrac {x ^ 2} {a 4T} + \ dfrac {Y ^ 2} {3T} =. 1 \) , where \ (T> 0 \) . the \ (C \) point is \ (G \) point source line on the ellipse \ [x \ = 3t] on while \ (C (3,0) \) , so \ (. 1 = T \) , then the request can be obtained for the elliptic equation \ (\ dfrac {x ^ 2 } {4} + \ dfrac Y ^ {2} = {}. 3. 1 \) .
method two
Disposed to be demand elliptic equation \ (\ dfrac {X ^ 2} {a 4T ^ 2} + \ dfrac {Y ^ 2} {3T ^ 2} =. 1 \) , where \ (T> 0 \) , then \ ( A, B, E \) three coordinates may be referred to as
\ [\ begin {split} & A \ left (2t \ cos \ alpha, - \ sqrt {3} t \ sin \ alpha \ right), \\ & B \ left (2t \ cos \ alpha, \ sqrt {3} t \ sin \ alpha \ right), \\ & E \ left (2t \ cos \ beta, \ sqrt {3} t \ sin \ beta \ right). \ end { split} \]
Since \ (the bE \) line \ (X \) intercept is \ (3 \) , and therefore can be obtained by the intercept coordinates of the formula
\ [\ begin {split} 3 & = \ dfrac {2t \ cos \ alpha \ cdot \ sqrt {3} t \ sin \ beta-2t \ cos \ beta \ cdot \ sqrt {3} t \ sin \ alpha} {\ sqrt {3} t \ sin \ beta- \ sqrt {3} t \ sin \ alpha} \\ & = \ dfrac {2t \ sin \ left (\ beta- \ alpha \ right)} {\ sin \ beta- \ sin \ alpha} \\ & = \ dfrac {2t \ cos \ dfrac {\ beta- \ alpha} {2
}} {\ cos \ dfrac {\ beta + \ alpha} {2}}. \ end {split} \] Similarly, by a straight line \ (the AE \) The \ (X \) intercept is \ (\ dfrac {. 4} {. 3} \) , available \ [\ dfrac {4} { 3} = \ dfrac {2t \ cos \ dfrac {\ beta + \ alpha} { 2}} {\ cos \ dfrac {\ beta- \ alpha} {2}}. \] the above two equations can be obtained by multiplying $
. 4 = ^ 2 $ a 4T, thereby to obtain solution \ (. 1 = T \) , then the elliptic equation is required \ (\ dfrac X ^ {2}. 4} + {\ dfrac Y ^ {2} = {}. 3. 1 \) .