Answer to Question #120855 in Analytic Geometry for Mst. Bithy

Question

Determine the equation of the hyperbola

Answer 1

$\text{Standard Form of the Equation of a Hyperbola}\\[1 em] \text{A hyperbola is the set of all points such that the difference of the distances between}\\[1 em] \text{ any point on the hyperbola and two fixed points is constant. The two fixed points }\\[1 em] \text{ are called the foci of the hyperbola.}\\[1 em] \text{We place the foci} F_1, F_2 \\[1 em] \text{ The center of this hyperbola is c(h,k).}\\[1 em] \text{ We let p(x,y)represent the coordinates of any point, on the hyperbola.}\\[1 em] \text{What does the definition of a hyperbola} \\[1 em] \text{The absolute value of the difference of the distances from the two foci} \\[1 em] \left |P F_{1}-P F_{2} \right | \text{must be constant. We denote this constant by 2a} \\[1 em] \text{Use the distance formula.} \\[1 em] P F_{1}-P F_{2} =2 a\\[1 em] \sqrt{[x-(h-c)]^{2}+(y-k)^{2}}-\sqrt{[x-(h+c)]^{2}+(y-k)^{2}} =2 a\\[1 em] \sqrt{[(x-h)+c]^{2}+(y-k)^{2}}-\sqrt{[(x-h)-c]^{2}+(y-k)^{2}} =2 a \\[1 em] \sqrt{[(x-h)+c]^{2}+(y-k)^{2}} =2 a+\sqrt{[(x-h)-c]^{2}+(y-k)^{2}} \\[1 em] \text{Squaring the sides} \\[1 em] (x-h)^{2}+2 c(x-h)+c^{2}+(y-k)^{2}\\[1 em] =4 a^{2}+4 a \sqrt{[(x-h)-c]^{2}+(y-k)^{2}}+(x-h)^{2}-2 c(x-h)+c^{2}+(y-k)^{2} \\[1 em] -4 a \sqrt{[(x-h)-c]^{2}+(y-k)^{2}} =4 a^{2}-4 c(x-h) \\[1 em] a \sqrt{[(x-h)-c]^{2}+(y-k)^{2}} =-a^{2}+c(x-h)\\[1 em] a^{2}\left[(x-h)^{2}-2 c(x-h)+c^{2}+(y-k)^{2}\right] =a^{4}-2 a^{2} c(x-h)+c^{2}(x-h)^{2} \\[1 em] a^{2}(x-h)^{2}-2 a^{2} c(x-h)+a^{2} c^{2}+a^{2}(y-k)^{2} =a^{4}-2 a^{2} c(x-h)+c^{2}(x-h)^{2} \\[1 em] a^{2}(x-h)^{2}-c^{2}(x-h)^{2}+a^{2}(y-k)^{2} =a^{4}-a^{2} c^{2} \\[1 em] \left(a^{2}-c^{2}\right)(x-h)^{2}+a^{2}(y-k)^{2} =a^{2}\left(a^{2}-c^{2}\right) \\[1 em] -b^{2}(x-h)^{2}+a^{2}(y-k)^{2} =a^{2}\left(-b^{2}\right) \\[1 em] \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}} =1$