Question Statement For any point on a hyperbola, the difference of its distances from two fixed points ( 2 , 2 ) (2, 2) ( 2 , 2 ) ( 10 , 2 ) (10, 2) ( 10 , 2 )
Background and Explanation A hyperbola is defined as the locus of points where the absolute difference of distances from two fixed points (called foci) is constant. This property is mathematically represented as:
∣ P F − P F ′ ∣ = 2 a , |PF - PF'| = 2a, ∣ PF − P F ′ ∣ = 2 a , where P ( x , y ) P(x, y) P ( x , y ) F F F F ′ F' F ′ 2 a 2a 2 a
In this problem, the foci are located at ( 2 , 2 ) (2, 2) ( 2 , 2 ) ( 10 , 2 ) (10, 2) ( 10 , 2 )
Solution Step 1: Write the hyperbolic condition
The given property of the hyperbola is:
( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 = 6. \sqrt{(x-10)^2 + (y-2)^2} - \sqrt{(x-2)^2 + (y-2)^2} = 6. ( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 = 6. Step 2: Square both sides to eliminate the square root
Squaring the equation:
( ( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 ) 2 = 36. \left(\sqrt{(x-10)^2 + (y-2)^2} - \sqrt{(x-2)^2 + (y-2)^2}\right)^2 = 36. ( ( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 ) 2 = 36. Expanding:
( x − 10 ) 2 + ( y − 2 ) 2 + ( x − 2 ) 2 + ( y − 2 ) 2 − 2 ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 36. (x-10)^2 + (y-2)^2 + (x-2)^2 + (y-2)^2 - 2\sqrt{\left((x-10)^2 + (y-2)^2\right)\left((x-2)^2 + (y-2)^2\right)} = 36. ( x − 10 ) 2 + ( y − 2 ) 2 + ( x − 2 ) 2 + ( y − 2 ) 2 − 2 ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 36. Simplify:
2 x 2 + 2 y 2 − 24 x − 8 y + 112 − 2 ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 36. 2x^2 + 2y^2 - 24x - 8y + 112 - 2\sqrt{\left((x-10)^2 + (y-2)^2\right)\left((x-2)^2 + (y-2)^2\right)} = 36. 2 x 2 + 2 y 2 − 24 x − 8 y + 112 − 2 ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 36. Rearranging:
x 2 + y 2 − 12 x − 4 y + 56 − ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 18. x^2 + y^2 - 12x - 4y + 56 - \sqrt{\left((x-10)^2 + (y-2)^2\right)\left((x-2)^2 + (y-2)^2\right)} = 18. x 2 + y 2 − 12 x − 4 y + 56 − ( ( x − 10 ) 2 + ( y − 2 ) 2 ) ( ( x − 2 ) 2 + ( y − 2 ) 2 ) = 18. Step 3: Substitute a variable and simplify further
Let:
t = x 2 + y 2 − 4 x − 4 y + 8. t = x^2 + y^2 - 4x - 4y + 8. t = x 2 + y 2 − 4 x − 4 y + 8. Substituting t t t
( t − 16 x + 96 ) t = t − 8 x + 30. \sqrt{(t - 16x + 96)t} = t - 8x + 30. ( t − 16 x + 96 ) t = t − 8 x + 30. Square both sides again:
( t − 16 x + 96 ) ( t ) = t 2 + 64 x 2 − 16 x t + 60 t + 900. (t - 16x + 96)(t) = t^2 + 64x^2 - 16xt + 60t + 900. ( t − 16 x + 96 ) ( t ) = t 2 + 64 x 2 − 16 x t + 60 t + 900. Expand and simplify:
t 2 − 16 x t + 96 t = t 2 + 64 x 2 − 16 x t + 60 t + 900 − 480 x . t^2 - 16xt + 96t = t^2 + 64x^2 - 16xt + 60t + 900 - 480x. t 2 − 16 x t + 96 t = t 2 + 64 x 2 − 16 x t + 60 t + 900 − 480 x . Cancel terms:
− 96 t + 64 x 2 + 60 t + 900 − 480 x = 0. -96t + 64x^2 + 60t + 900 - 480x = 0. − 96 t + 64 x 2 + 60 t + 900 − 480 x = 0. Divide by 4:
16 x 2 − 120 x − 9 t + 225 = 0. 16x^2 - 120x - 9t + 225 = 0. 16 x 2 − 120 x − 9 t + 225 = 0. Step 4: Replace t t t
Substitute back t = x 2 + y 2 − 4 x − 4 y + 8 t = x^2 + y^2 - 4x - 4y + 8 t = x 2 + y 2 − 4 x − 4 y + 8
16 x 2 − 120 x − 9 ( x 2 + y 2 − 4 x − 4 y + 8 ) + 225 = 0. 16x^2 - 120x - 9(x^2 + y^2 - 4x - 4y + 8) + 225 = 0. 16 x 2 − 120 x − 9 ( x 2 + y 2 − 4 x − 4 y + 8 ) + 225 = 0. Simplify:
16 x 2 − 120 x − 9 x 2 − 9 y 2 + 36 x + 36 y − 72 + 225 = 0. 16x^2 - 120x - 9x^2 - 9y^2 + 36x + 36y - 72 + 225 = 0. 16 x 2 − 120 x − 9 x 2 − 9 y 2 + 36 x + 36 y − 72 + 225 = 0. Combine like terms:
7 x 2 − 9 y 2 − 84 x + 36 y + 153 = 0. 7x^2 - 9y^2 - 84x + 36y + 153 = 0. 7 x 2 − 9 y 2 − 84 x + 36 y + 153 = 0.
Hyperbolic property :
∣ P F − P F ′ ∣ = 2 a . |PF - PF'| = 2a. ∣ PF − P F ′ ∣ = 2 a .
Squaring technique : Eliminates square roots to simplify the equation.Variable substitution : Used to manage complexity and simplify terms.
Summary of Steps
Express the condition : Write the hyperbolic condition as:
( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 = 6. \sqrt{(x-10)^2 + (y-2)^2} - \sqrt{(x-2)^2 + (y-2)^2} = 6. ( x − 10 ) 2 + ( y − 2 ) 2 − ( x − 2 ) 2 + ( y − 2 ) 2 = 6.
Eliminate square roots : Square the equation twice to simplify.Simplify and substitute : Use substitution to manage terms and simplify the resulting equation.Final result : The equation of the hyperbola is:
7 x 2 − 9 y 2 − 84 x + 36 y + 153 = 0. 7x^2 - 9y^2 - 84x + 36y + 153 = 0. 7 x 2 − 9 y 2 − 84 x + 36 y + 153 = 0. 
