Stephanie Hunter

Stephanie Hunter

Answered

2022-07-17

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.

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Answer & Explanation

minotaurafe

minotaurafe

Expert

2022-07-18Added 22 answers

Step 1
Consider an ellipse with semimajor axis a and semiminor axis b centered at (0,0), where a b > 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 2
Using 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 .

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