Prove that the locus of the incenter of the Δ P S S ′ is an ellipse of eccentricity 2 e 1 + e Let S and S′ be the foci of an ellipse whose eccentricity is e.P is a variable point on the ellipse.Prove that the locus of the incenter of the Δ P S S ′ is an ellipse of eccentricity 2 e 1 + e .Let P be ( a cos ⁡ θ , b sin ⁡ θ ). Let the incenter of the triangle PSS′ be (h,k). The formula for the incenter of a triangle whose side lengths are a,b,c and whose vertices have coordinates ( x 1 , y 1 ) , ( x 2 , y 2 ) , and ( x 3 , y 3 ) is ( a x 1 + b x 2 + c x 3 a + b + c , a y 1 + b y 2 + c y 3 a + b + c ) .Then, h = 2 c ⋅ a cos ⁡ θ + P S ′ ⋅ c − P S ⋅ c 2 c + P S ′ + P S and k = 2 c ⋅ a sin ⁡ θ 2 c + P S ′ + P S ,but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

Question

Prove that the locus of the incenter of the   Δ  P  S      S    ′   is an ellipse of eccentricity                     2        e                    1        +        e            Let S and S′ be the foci of an ellipse whose eccentricity is e.P is a variable point on the ellipse.Prove that the locus of the incenter of the   Δ  P  S      S    ′   is an ellipse of eccentricity                     2        e                    1        +        e            .Let P be   (  a  cos  ⁡  θ  ,  b  sin  ⁡  θ  ). Let the incenter of the triangle PSS′ be (h,k). The formula for the incenter of a triangle whose side lengths are a,b,c and whose vertices have coordinates       (          x      1        ,          y      1        )  ,       (          x      2        ,          y      2        )  , and       (          x      3        ,          y      3        )   is      (                  a                  x          1                +        b                  x          2                +        c                  x          3                            a        +        b        +        c              ,                  a                  y          1                +        b                  y          2                +        c                  y          3                            a        +        b        +        c              )      .Then,   h  =            2      c      ⋅      a      cos      ⁡      θ      +      P              S        ′            ⋅      c      −      P      S      ⋅      c              2      c      +      P              S        ′            +      P      S         and   k  =            2      c      ⋅      a      sin      ⁡      θ              2      c      +      P              S        ′            +      P      S          ,but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

minotaurafe · Accepted Answer

Step 1Consider an ellipse with semimajor axis a and semiminor axis b centered at (0,0), where   a  ≥  b  &amp;gt;  0. The eccentricity e of this ellipse is given by   e  =      1    −                  b        2                    a        2            . Without loss of generality, let   S  =  (  +  c  ,  0  ) and       S    ′    =  (  −  c  ,  0  ) be the foci of this ellipse, where   c  :=  a  e. Note that   P  S  +  P      S    ′    =  2  a for every point P on the ellipse. Thus, if             (        h  (  θ  )  ,  k  (  θ  )            )       is the incenter of   P  =            (        a    cos  ⁡  (  θ  )  ,  b    sin  ⁡  (  θ  )            )       for   θ  ∈  [  0  ,  2  π  ), then, as you have found out,  h  (  θ  )  =            2              a        2            e            cos      ⁡      (      θ      )      −      a      e            P      S      +      a      e            P              S        ′                    2      a      e      +      2      a        =      e          e      +      1                  (        a  cos  ⁡  (  θ  )  −      1    2    P  S  +      1    2    P      S    ′              )          ..and   k  (  θ  )  =            2      a      b      e            sin      ⁡      (      θ      )              2      a      e      +      2      a        =      e          e      +      1                  (        b    sin  ⁡  (  θ  )            )          ..Step 2Using   e  =      1    −                  b        2                    a        2             with   P  S  =                    (              a        cos    ⁡    (    θ    )    −    a    e          )      2        +          b      2              sin      2        ⁡    (    θ    )         and   P      S    ′    =                    (              a        cos    ⁡    (    θ    )    +    a    e          )      2        +          b      2              sin      2        ⁡    (    θ    )      ,it can be easily seen that   P  S  =  a            (        1  −  e    cos  ⁡  (  θ  )            )       and   P      S    ′    =  a            (        1  +  e    cos  ⁡  (  θ  )            )      . Therefore,  h  (  θ  )  =      e          e      +      1        a  (  1  +  e  )    cos  ⁡  (  θ  )  =  a  e    cos  ⁡  (  θ  )    .Let   A  :=  a  e and   B  :=      e          e      +      1        b. Then,  h  (  θ  )  =  A    cos  ⁡  (  θ  )   and   k  (  θ  )  =  B    sin  ⁡  (  θ  )    .Therefore, the locus of the incenter of PSS′ is indeed an ellipse with the semimajor axis A and the semiminor axis B. If E is its eccentricity, then   E  =      1    −                  B        2                    A        2              =      1    −                            b          2                          /                          a          2                            (        e        +        1                  )          2                      =      1    −                  1        −                  e          2                            (        e        +        1                  )          2                      =      1    −                  1        −        e                    e        +        1              =                    2        e                    e        +        1                .

Prove that the locus of the incenter of the triangle PSS′ is an ellipse of eccentricity sqrt{(2e)/(1+e)}

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