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Finding Equation of Hyperbola from given 2 Foci and Transverse Axis Length

1. Given a Hyperbola having Coordinates of Center $(x_c,y_c)$, Coordinates of 2 Foci $(x_{f1},y_{f1})$ and $(x_{f2},y_{f2})$ and Length of Transverse Axis $L=2a$, the Equation of the Hyperbola can be found out as follows
   
   As per definition of Hyperbola, the Difference of the Distance from 2 Foci to Any Point $(x,y)$ on the Hyperbola is equal to the Length of it's Transverse Axis. Hence
   
   $\sqrt{{(x-x_{f1})}^2 + {(y-y_{f1})}^2} - \sqrt{{(x-x_{f2})}^2 + {(y-y_{f2})}^2} =L=2a$
   
   $\Rightarrow \sqrt{{(x-x_{f1})}^2 + {(y-y_{f1})}^2} = 2a + \sqrt{{(x-x_{f2})}^2 + {(y-y_{f2})}^2}$   ...(1)
   
   Squaring Both Sides of equation (1) above we get
   
   ${(x-x_{f1})}^2 + {(y-y_{f1})}^2 = 4a^2 + {(x-x_{f2})}^2 + {(y-y_{f2})}^2 + 4a \sqrt{{(x-x_{f2})}^2 + {(y-y_{f2})}^2}$
   
   $\Rightarrow 4a^2 + {(x-x_{f2})}^2 + {(y-y_{f2})}^2 - {(x-x_{f1})}^2 - {(y-y_{f1})}^2 = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_{f2})}^2}$   ...(2)
   
   Squaring Both Sides of equation (2) above we get
   
   $\Rightarrow {(4a^2 + {(x-x_{f2})}^2 + {(y-y_{f2})}^2 - {(x-x_{f1})}^2 - {(y-y_{f1})}^2)}^2 = 16a^2 ({(x-x_{f2})}^2 + {(y-y_{f2})}^2)$
   
   $\Rightarrow {(4a^2 + {(x-x_{f2})}^2 + {(y-y_{f2})}^2 - {(x-x_{f1})}^2 - {(y-y_{f1})}^2)}^2 -16a^2 ({(x-x_{f2})}^2 + {(y-y_{f2})}^2)=0$   ...(3)
   
   Expanding and Simplifying equation (3) above we get
   
   $(4{x_{f1}}^2+4{x_{f2}}^2-8{x_{f1}}{x_{f2}}-16a^2)x^2 + (4{y_{f1}}^2+4{y_{f2}}^2-8{y_{f1}}{y_{f2}}-16a^2) y^2 +(8{x_{f1}}{y_{f1}}-8{x_{f1}}{y_{f2}}+8{x_{f2}}{y_{f2}}-8{y_{f1}}{x_{f2}})xy$
   $+ (16a^2{x_{f1}}+16a^2{x_{f2}}-4{x_{f1}}^3-4{x_{f2}}^3+4{x_{f1}}{x_{f2}}^2+4{x_{f1}}{y_{f2}}^2-4{x_{f1}}{y_{f1}}^2+4{x_{f2}}{x_{f1}}^2+4{x_{f2}}{y_{f1}}^2-4{x_{f2}}{y_{f2}}^2)x$
   $+ (16a^2{y_{f1}}+16a^2{y_{f2}}-4{y_{f1}}^3-4{y_{f2}}^3+4{y_{f1}}{y_{f2}}^2+4{y_{f1}}{x_{f2}}^2-4{y_{f1}}{x_{f1}}^2+4{y_{f2}}{y_{f1}}^2+4{y_{f2}}{x_{f1}}^2-4{y_{f2}}{x_{f2}}^2)y$
   $+ (16a^4+{x_{f1}}^4+{y_{f1}}^4+{x_{f2}}^4+{y_{f2}}^4-8a^2{x_{f1}}^2-8a^2{y_{f1}}^2-8a^2{x_{f2}}^2-8a^2{y_{f2}}^2+2{x_{f1}}^2{y_{f1}}^2+2{x_{f2}}^2{y_{f2}}^2-2{x_{f1}}^2{x_{f2}}^2-2{y_{f1}}^2{x_{f2}}^2-2{x_{f1}}^2{y_{f2}}^2-2{y_{f1}}^2{y_{f2}}^2)=0$   ...(4)
   
   The equation (4) above gives the Equation of Hyperbola having Coordinates of 2 Foci at $(x_{f1},y_{f1})$ and $(x_{f2},y_{f2})$ and Length of Transverse Axis $L=2a$ if $L < F$, where $F$ is Distance Between 2 Foci.
   
   Please note that the equation (4) above can also be used for Finding Equation of Ellipse from given 2 Foci and Major Axis Length.
2. The Hyperbola represented by equation (4) gets converted to an $X$-Transverse Hyperbola on setting $y_{f1}=y_{f2}=y_f$ as given by the following calculations
   
   $(4{x_{f1}}^2+4{x_{f2}}^2-8{x_{f1}}{x_{f2}}-16a^2)x^2 + (4{y_f}^2+4{y_f}^2-8{y_f}{y_f}-16a^2) y^2 +(8{x_{f1}}{y_f}-8{x_{f1}}{y_f}+8{x_{f2}}{y_f}-8{y_f}{x_{f2}})xy$
   $+ (16a^2{x_{f1}}+16a^2{x_{f2}}-4{x_{f1}}^3-4{x_{f2}}^3+4{x_{f1}}{x_{f2}}^2+4{x_{f1}}{y_f}^2-4{x_{f1}}{y_f}^2+4{x_{f2}}{x_{f1}}^2+4{x_{f2}}{y_f}^2-4{x_{f2}}{y_f}^2)x$
   $+ (16a^2{y_f}+16a^2{y_f}-4{y_f}^3-4{y_f}^3+4{y_f}{y_f}^2+4{y_f}{x_{f2}}^2-4{y_f}{x_{f1}}^2+4{y_f}{y_f}^2+4{y_f}{x_{f1}}^2-4{y_f}{x_{f2}}^2)y$
   $+ (16a^4+{x_{f1}}^4+{y_f}^4+{x_{f2}}^4+{y_f}^4-8a^2{x_{f1}}^2-8a^2{y_f}^2-8a^2{x_{f2}}^2-8a^2{y_f}^2+2{x_{f1}}^2{y_f}^2+2{x_{f2}}^2{y_f}^2-2{x_{f1}}^2{x_{f2}}^2-2{y_f}^2{x_{f2}}^2-2{x_{f1}}^2{y_f}^2-2{y_f}^2{y_f}^2)=0$
   
   $\Rightarrow (4({x_{f1}}^2+{x_{f2}}^2-2{x_{f1}}{x_{f2}})-16a^2)x^2 + (8{y_f}^2-8{y_f}^2-16a^2) y^2 + (16a^2{x_{f1}}+16a^2{x_{f2}}-4{x_{f1}}^3-4{x_{f2}}^3+4{x_{f1}}{x_{f2}}^2+4{x_{f2}}{x_{f1}}^2)x$
   $+ (32a^2{y_f}-4{y_f}^3-4{y_f}^3+4{y_f}^3+4{y_f}^3)y +(16a^4+{x_{f1}}^4+{x_{f2}}^4-2{x_{f1}}^2{x_{f2}}^2+2{y_f}^4-2{y_f}^4-16a^2{y_f}^2-8a^2({x_{f1}}^2+{x_{f2}}^2))=0$
   
   $\Rightarrow (4{(x_{f1}-x_{f2})}^2-16a^2)x^2 -16a^2y^2 + (16a^2(x_{f1}+x_{f2})-4x_{f1}({x_{f1}}^2-{x_{f2}}^2) +4x_{f2}({x_{f1}}^2-{x_{f2}}^2))x$
   $+ 32a^2y_fy +(16a^4+{({x_{f1}}^2-{x_{f2}}^2)}^2-16a^2{y_f}^2-8a^2({x_{f1}}^2+{x_{f2}}^2))=0$
   
   $\Rightarrow (4{(x_{f1}-x_{f2})}^2-16a^2)x^2 -16a^2y^2 + (16a^2(x_{f1}+x_{f2})-4x_{f1}(x_{f1}-x_{f2})(x_{f1}+x_{f2}) +4x_{f2}(x_{f1}-x_{f2})(x_{f1}+x_{f2}))x$
   $+ 32a^2y_fy +(16a^4+{((x_{f1}-x_{f2})(x_{f1}+x_{f2}))}^2-16a^2{y_f}^2-8a^2({x_{f1}}^2+{x_{f2}}^2))=0$   ...(5)
   
   Now, we know that
   
   $x_{f1}-x_{f2}=2c$   ...(6)
   
   $\Rightarrow {(x_{f1}-x_{f2})}^2=4c^2 \hspace{.5cm}\Rightarrow {x_{f1}}^2+{x_{f2}}^2-2{x_{f1}}{x_{f2}}=4c^2 \hspace{.5cm}\Rightarrow {x_{f1}}^2+{x_{f2}}^2=4c^2+2{x_{f1}}{x_{f2}}$   ...(7)
   
   Also
   
   $\frac{x_{f1}+x_{f2}}{2}=x_c\hspace{.5cm}\Rightarrow x_{f1}+x_{f2}=2x_c$   (where $(x_c,y_c)$ are Coordinates of Center of Hyperbola) ...(8)
   
   $\Rightarrow {(x_{f1}+x_{f2})}^2=4{x_c}^2 \hspace{.5cm}\Rightarrow {x_{f1}}^2+{x_{f2}}^2+2{x_{f1}}{x_{f2}}=4{x_c}^2 \hspace{.5cm}\Rightarrow {x_{f1}}^2+{x_{f2}}^2=4{x_c}^2-2{x_{f1}}{x_{f2}}$   ...(9)
   
   Adding equations (7) and (9) we get
   
   $2({x_{f1}}^2+{x_{f2}}^2)=4c^2+2{x_{f1}}{x_{f2}}+4{x_c}^2-2{x_{f1}}{x_{f2}}\hspace{.5cm}\Rightarrow 2({x_{f1}}^2+{x_{f2}}^2)=4(c^2+{x_c}^2)\hspace{.5cm}\Rightarrow ({x_{f1}}^2+{x_{f2}}^2)=2(c^2+{x_c}^2)$   ...(10)
   
   Now, putting the values of $x_{f1}-x_{f2}$, $x_{f1}+x_{f2}$ and ${x_{f1}}^2+{x_{f2}}^2$ from equations (6), (8) and (10) in equation (5) we get
   
   $(4{(2c)}^2-16a^2)x^2 -16a^2y^2 + (16a^2(2x_c)-4x_{f1}(2c)(2x_c) +4x_{f2}(2c)(2x_c))x+ 32a^2y_fy +(16a^4+{((2c)(2x_c))}^2-16a^2{y_f}^2-8a^2(2(c^2+{x_c}^2)))=0$
   
   $\Rightarrow (16c^2-16a^2)x^2 -16a^2y^2 + (32a^2x_c -16x_{f1}cx_c +16x_{f2}cx_c)x + 32a^2y_fy +(16a^4+16c^2{x_c}^2-16a^2{y_f}^2-16a^2{x_c}^2-16a^2c^2)=0$
   
   $\Rightarrow -16(a^2-c^2)x^2 -16a^2y^2 + (32a^2x_c -16cx_c(x_{f1} -x_{f2}))x + 32a^2y_fy +(16a^4-16a^2c^2+16c^2{x_c}^2-16a^2{x_c}^2-16a^2{y_f}^2)=0$
   
   $\Rightarrow -16(a^2-c^2)x^2 -16a^2y^2 + (32a^2x_c -16cx_c(2c))x + 32a^2y_fy +(16a^2(a^2-c^2) - 16{x_c}^2(a^2-c^2) -16a^2{y_f}^2)=0$
   
   $\Rightarrow -16(a^2-c^2)x^2 -16a^2y^2 + (32a^2x_c -32c^2x_c)x + 32a^2y_fy +(16a^2(a^2-c^2) - 16{x_c}^2(a^2-c^2) -16a^2{y_f}^2)=0$
   
   $\Rightarrow -16(a^2-c^2)x^2 -16a^2y^2 + 32x_c (a^2 -c^2)x + 32a^2y_fy +(16a^2(a^2-c^2) - 16{x_c}^2(a^2-c^2) -16a^2{y_f}^2)=0$   ...(11)
   
   Now, for any Hyperbola $a^2-c^2=-b^2$. Substituting it in equation (11) above we get
   
   $16b^2x^2 -16a^2y^2 - 32b^2x_cx + 32a^2y_fy - 16a^2b^2 + 16b^2{x_c}^2 -16a^2{y_f}^2=0$   ...(12)
   
   Dividing the equation (12) above by 16 on Both Sides we get
   
   $b^2x^2 - a^2y^2 - 2b^2x_cx + 2a^2y_fy + b^2{x_c}^2 - a^2{y_f}^2 - a^2b^2=0$   ...(13)
   
   Also for $X$-Transverse Hyperbola $y_f=y_c$. Hence,
   
   $b^2x^2 - a^2y^2 - 2b^2x_cx + 2a^2y_cy + b^2{x_c}^2 - a^2{y_c}^2 - a^2b^2=0$   ...(14)
   
   The equation (14) above gives the equation of $X$-Transverse Hyperbola.
3. The Hyperbola represented by equation (4) gets converted to an $Y$-Transverse Hyperbola on setting $x_{f1}=x_{f2}=x_f$ as given by the following calculations
   
   $(4{x_f}^2+4{x_f}^2-8{x_f}{x_f}-16a^2)x^2 + (4{y_{f1}}^2+4{y_{f2}}^2-8{y_{f1}}{y_{f2}}-16a^2) y^2 +(8{x_f}{y_{f1}}-8{x_f}{y_{f2}}+8{x_f}{y_{f2}}-8{y_{f1}}{x_f})xy$
   $+ (16a^2{x_f}+16a^2{x_f}-4{x_f}^3-4{x_f}^3+4{x_f}{x_f}^2+4{x_f}{y_{f2}}^2-4{x_f}{y_{f1}}^2+4{x_f}{x_f}^2+4{x_f}{y_{f1}}^2-4{x_f}{y_{f2}}^2)x$
   $+ (16a^2{y_{f1}}+16a^2{y_{f2}}-4{y_{f1}}^3-4{y_{f2}}^3+4{y_{f1}}{y_{f2}}^2+4{y_{f1}}{x_f}^2-4{y_{f1}}{x_f}^2+4{y_{f2}}{y_{f1}}^2+4{y_{f2}}{x_f}^2-4{y_{f2}}{x_f}^2)y$
   $+ (16a^4+{x_f}^4+{y_{f1}}^4+{x_f}^4+{y_{f2}}^4-8a^2{x_f}^2-8a^2{y_{f1}}^2-8a^2{x_f}^2-8a^2{y_{f2}}^2+2{x_f}^2{y_{f1}}^2+2{x_f}^2{y_{f2}}^2-2{x_f}^2{x_f}^2-2{y_{f1}}^2{x_f}^2-2{x_f}^2{y_{f2}}^2-2{y_{f1}}^2{y_{f2}}^2)=0$   ...(1)
   
   $\Rightarrow (8{x_f}^2-8{x_f}^2-16a^2) x^2 + (4({y_{f1}}^2+{y_{f2}}^2-2{y_{f1}}{y_{f2}})-16a^2)y^2 + (32a^2{x_f}-4{x_f}^3-4{x_f}^3+4{x_f}^3+4{x_f}^3)x$
   $+ (16a^2{y_{f1}}+16a^2{y_{f2}}-4{y_{f1}}^3-4{y_{f2}}^3+4{y_{f1}}{y_{f2}}^2+4{y_{f2}}{y_{f1}}^2)y +(16a^4+{y_{f1}}^4+{y_{f2}}^4-2{y_{f1}}^2{y_{f2}}^2+2{x_f}^4-2{x_f}^4-16a^2{x_f}^2-8a^2({y_{f1}}^2+{y_{f2}}^2))=0$
   
   $\Rightarrow -16a^2x^2 -(4{(y_{f1}-y_{f2})}^2-16a^2)y^2 + 32a^2x_fx + (16a^2(y_{f1}+y_{f2})-4y_{f1}({y_{f1}}^2-{y_{f2}}^2) +4y_{f2}({y_{f1}}^2-{y_{f2}}^2))y$
   $+(16a^4+{({y_{f1}}^2-{y_{f2}}^2)}^2-16a^2{x_f}^2-8a^2({y_{f1}}^2+{y_{f2}}^2))=0$
   
   $\Rightarrow -16a^2x^2 + (4{(y_{f1}-y_{f2})}^2-16a^2)y^2 + 32a^2x_fx + (16a^2(y_{f1}+y_{f2})-4y_{f1}(y_{f1}-y_{f2})(y_{f1}+y_{f2}) +4y_{f2}(y_{f1}-y_{f2})(y_{f1}+y_{f2}))y$
   $+(16a^4+{((y_{f1}-y_{f2})(y_{f1}+y_{f2}))}^2-16a^2{x_f}^2-8a^2({y_{f1}}^2+{y_{f2}}^2))=0$   ...(15)
   
   Now, we know that
   
   $y_{f1}-y_{f2}=2c$   ...(16)
   
   $\Rightarrow {(y_{f1}-y_{f2})}^2=4c^2 \hspace{.5cm}\Rightarrow {y_{f1}}^2+{y_{f2}}^2-2{y_{f1}}{y_{f2}}=4c^2 \hspace{.5cm}\Rightarrow {y_{f1}}^2+{y_{f2}}^2=4c^2+2{y_{f1}}{y_{f2}}$   ...(17)
   
   Also
   
   $\frac{y_{f1}+y_{f2}}{2}=y_c\hspace{.5cm}\Rightarrow y_{f1}+y_{f2}=2y_c$   (where $(x_c,y_c)$ are Coordinates of Center of Hyperbola) ...(18)
   
   $\Rightarrow {(y_{f1}+y_{f2})}^2=4{y_c}^2 \hspace{.5cm}\Rightarrow {y_{f1}}^2+{y_{f2}}^2+2{y_{f1}}{y_{f2}}=4{y_c}^2 \hspace{.5cm}\Rightarrow {y_{f1}}^2+{y_{f2}}^2=4{y_c}^2-2{y_{f1}}{y_{f2}}$   ...(19)
   
   Adding equations (17) and (19) we get
   
   $2({y_{f1}}^2+{y_{f2}}^2)=4c^2+2{y_{f1}}{y_{f2}}+4{y_c}^2-2{y_{f1}}{y_{f2}}\hspace{.5cm}\Rightarrow 2({y_{f1}}^2+{y_{f2}}^2)=4(c^2+{y_c}^2)\hspace{.5cm}\Rightarrow ({y_{f1}}^2+{y_{f2}}^2)=2(c^2+{y_c}^2)$   ...(20)
   
   Now, putting the values of $y_{f1}-y_{f2}$, $y_{f1}+y_{f2}$ and ${y_{f1}}^2+{y_{f2}}^2$ from equations (16), (18) and (20) in equation (15) we get
   
   $-16a^2x^2 + (4{(2c)}^2-16a^2)y^2 + 32a^2x_fx + (16a^2(2y_c)-4y_{f1}(2c)(2y_c) +4y_{f2}(2c)(2y_c))x +(16a^4+{((2c)(2y_c))}^2-16a^2{x_f}^2-8a^2(2(c^2+{y_c}^2)))=0$
   
   $\Rightarrow -16a^2x^2 +(16c^2-16a^2)y^2 + 32a^2x_fx + (32a^2y_c -16y_{f1}cy_c +16y_{f2}cy_c)y +(16a^4+16c^2{y_c}^2-16a^2{x_f}^2-16a^2{y_c}^2-16a^2c^2)=0$
   
   $\Rightarrow -16a^2x^2 -16(a^2-c^2)y^2 + 32a^2x_fx + (32a^2y_c -16cy_c(y_{f1} -y_{f2}))y +(16a^4-16a^2c^2+16c^2{y_c}^2-16a^2{y_c}^2-16a^2{x_f}^2)=0$
   
   $\Rightarrow -16a^2x^2 -16(a^2-c^2)y^2 + 32a^2x_fx + (32a^2y_c -16cy_c(2c))y +(16a^2(a^2-c^2) - 16{y_c}^2(a^2-c^2) -16a^2{x_f}^2)=0$
   
   $\Rightarrow -16a^2x^2 -16(a^2-c^2)y^2 + 32a^2x_fx + (32a^2y_c -32c^2y_c)y +(16a^2(a^2-c^2) - 16{y_c}^2(a^2-c^2) -16a^2{x_f}^2)=0$
   
   $\Rightarrow -16a^2x^2 -16(a^2-c^2)y^2 + 32a^2x_fx + 32y_c (a^2 -c^2)y + (16a^2(a^2-c^2) - 16{y_c}^2(a^2-c^2) -16a^2{x_f}^2)=0$   ...(21)
   
   Now, for any Hyperbola $a^2-c^2=-b^2$. Substituting it in equation (21) above we get
   
   $-16a^2x^2 +16b^2y^2 + 32a^2x_fx - 32b^2y_cy - 16a^2b^2 + 16b^2{y_c}^2 -16a^2{x_f}^2=0$   ...(22)
   
   Dividing the equation (22) above by 16 on Both Sides we get
   
   $-a^2x^2 + b^2y^2 + 2a^2x_fx - 2b^2y_cy - a^2{x_f}^2 + b^2{y_c}^2 - a^2b^2=0$   ...(23)
   
   Also for $Y$-Transverse Hyperbola $x_f=x_c$. Hence,
   
   $-a^2x^2 + b^2y^2 + 2a^2x_cx - 2b^2y_cy - a^2{x_c}^2 + b^2{y_c}^2 - a^2b^2=0$   ...(24)
   
   The equation (24) above gives the equation of $Y$-Transverse Hyperbola.