Question Statement Find the equation of the ellipse as the locus of points P ( x , y ) P(x, y) P ( x , y ) P P P ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 1 , 1 ) (1, 1) ( 1 , 1 )
Background and Explanation An ellipse is the locus of points such that the sum of distances from any point on the ellipse to two fixed points (foci) is constant. This property allows us to derive the equation of the ellipse based on the given condition.
In this case:
The fixed points (foci) are ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 1 , 1 ) (1, 1) ( 1 , 1 )
The constant sum of distances is 2 2 2
The general approach is to set up the distance condition, simplify it using algebra, and derive the standard equation of the ellipse.
Solution We start by expressing the condition mathematically:
( x − 0 ) 2 + ( y − 0 ) 2 + ( x − 1 ) 2 + ( y − 1 ) 2 = 2 \sqrt{(x - 0)^2 + (y - 0)^2} + \sqrt{(x - 1)^2 + (y - 1)^2} = 2 ( x − 0 ) 2 + ( y − 0 ) 2 + ( x − 1 ) 2 + ( y − 1 ) 2 = 2 Step 1: Simplify the distance condition
Expanding the distances:
x 2 + y 2 + ( x − 1 ) 2 + ( y − 1 ) 2 = 2 \sqrt{x^2 + y^2} + \sqrt{(x - 1)^2 + (y - 1)^2} = 2 x 2 + y 2 + ( x − 1 ) 2 + ( y − 1 ) 2 = 2 Step 2: Square both sides
To eliminate the square roots:
x 2 + y 2 + x 2 − 2 x + 1 + y 2 − 2 y + 1 = 2 \sqrt{x^2 + y^2} + \sqrt{x^2 - 2x + 1 + y^2 - 2y + 1} = 2 x 2 + y 2 + x 2 − 2 x + 1 + y 2 − 2 y + 1 = 2 ( x 2 + y 2 ) + ( x 2 − 2 x + 1 + y 2 − 2 y + 1 ) + 2 ( x 2 + y 2 ) ( ( x 2 − 2 x + 1 ) + ( y 2 − 2 y + 1 ) ) = 4 (x^2 + y^2) + (x^2 - 2x + 1 + y^2 - 2y + 1) + 2\sqrt{(x^2 + y^2)((x^2 - 2x + 1) + (y^2 - 2y + 1))} = 4 ( x 2 + y 2 ) + ( x 2 − 2 x + 1 + y 2 − 2 y + 1 ) + 2 ( x 2 + y 2 ) (( x 2 − 2 x + 1 ) + ( y 2 − 2 y + 1 )) = 4 Simplify:
2 x 2 + 2 y 2 − 2 x − 2 y + 2 + 2 ( x 2 + y 2 ) ( x 2 + y 2 − 2 x − 2 y + 2 ) = 4 2x^2 + 2y^2 - 2x - 2y + 2 + 2\sqrt{(x^2 + y^2)(x^2 + y^2 - 2x - 2y + 2)} = 4 2 x 2 + 2 y 2 − 2 x − 2 y + 2 + 2 ( x 2 + y 2 ) ( x 2 + y 2 − 2 x − 2 y + 2 ) = 4 Step 3: Isolate the square root
Reorganize to isolate the square root term:
( x 2 + y 2 ) ( x 2 + y 2 − 2 x − 2 y + 2 ) = 2 − x 2 − y 2 + x + y − 1 \sqrt{(x^2 + y^2)(x^2 + y^2 - 2x - 2y + 2)} = 2 - x^2 - y^2 + x + y - 1 ( x 2 + y 2 ) ( x 2 + y 2 − 2 x − 2 y + 2 ) = 2 − x 2 − y 2 + x + y − 1 Let t = x 2 + y 2 t = x^2 + y^2 t = x 2 + y 2 t t t
t − x − y + 1 + t ( t − 2 x − 2 y + 2 ) = 2 t - x - y + 1 + \sqrt{t(t - 2x - 2y + 2)} = 2 t − x − y + 1 + t ( t − 2 x − 2 y + 2 ) = 2 Step 4: Square again
Square both sides again:
t 2 − 2 t x − 2 t y + 2 t = ( 1 − t + x + y ) 2 t^2 - 2tx - 2ty + 2t = (1 - t + x + y)^2 t 2 − 2 t x − 2 t y + 2 t = ( 1 − t + x + y ) 2 Expand:
t 2 − 2 t x − 2 t y + 2 t = 1 + t 2 − 2 t + x 2 + y 2 + 2 x y + 2 ( x + y − t x − t y ) t^2 - 2tx - 2ty + 2t = 1 + t^2 - 2t + x^2 + y^2 + 2xy + 2(x + y - tx - ty) t 2 − 2 t x − 2 t y + 2 t = 1 + t 2 − 2 t + x 2 + y 2 + 2 x y + 2 ( x + y − t x − t y ) Reorganize terms:
2 t = 1 − 2 t + x 2 + y 2 + 2 x y + 2 x + 2 y 2t = 1 - 2t + x^2 + y^2 + 2xy + 2x + 2y 2 t = 1 − 2 t + x 2 + y 2 + 2 x y + 2 x + 2 y Step 5: Replace t t t x 2 + y 2 x^2 + y^2 x 2 + y 2
Substitute t = x 2 + y 2 t = x^2 + y^2 t = x 2 + y 2
2 ( x 2 + y 2 ) = 1 − 2 ( x 2 + y 2 ) + x 2 + y 2 + 2 x y + 2 x + 2 y 2(x^2 + y^2) = 1 - 2(x^2 + y^2) + x^2 + y^2 + 2xy + 2x + 2y 2 ( x 2 + y 2 ) = 1 − 2 ( x 2 + y 2 ) + x 2 + y 2 + 2 x y + 2 x + 2 y Simplify:
3 x 2 + 3 y 2 − 2 x y − 2 x − 2 y = 0 3x^2 + 3y^2 - 2xy - 2x - 2y = 0 3 x 2 + 3 y 2 − 2 x y − 2 x − 2 y = 0 Step 6: Final Equation
Divide through by 3 to simplify:
x 2 + y 2 − 2 3 x y − 2 3 x − 2 3 y = 0 x^2 + y^2 - \frac{2}{3}xy - \frac{2}{3}x - \frac{2}{3}y = 0 x 2 + y 2 − 3 2 x y − 3 2 x − 3 2 y = 0 This is the equation of the ellipse in its implicit form.
Ellipse Definition : The sum of distances from any point on the ellipse to two fixed points is constant.Distance Formula :
Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Ellipse Equation Derivation : Using algebraic manipulation to simplify and isolate terms.
Summary of Steps
Express the condition ∣ PF ∣ + ∣ PF ′ ∣ = 2 |\text{PF}| + |\text{PF}'| = 2 ∣ PF ∣ + ∣ PF ′ ∣ = 2
Expand the distances and square both sides to eliminate square roots.
Simplify the resulting expressions using substitution.
Derive the standard implicit equation of the ellipse.
