Conjugate Hyperbola

1. For any given Hyperbola, it's corresponding Conjugate Hyperbola can be obtained by Rotating the Hyperbola by \(90^{\circ}\) (Clockwize or Counter Clockwize) and interchanging the Lengths of it's Transverse and Conjugate Axis.
2. As given in Derivation of Standard and Implicit Coordinate Equation for Axis Aligned Hyperbolas, the Standard Coordinate Equation for Hyperbolas having their Transverse Axes Parallel to the \(X\)-Axis, Center at Origin, Length of Semi-Transverse Axis \(a\) and Length of Semi-Conjugate Axis \(b\) is given as
   
   \(\frac{x^2}{a^2} - \frac{y^2}{b^2} =1 \)   ...(1)
   
   Rotating the equation (1) above Counter Clockwise by \(90^\circ\) we get
   
   \(\frac{{(x \cos 90^\circ + y \sin 90^\circ)} ^2}{a^2} - \frac{{(y \cos 90^\circ - x \sin 90^\circ)} ^2}{b^2} =1 \)
   
   \(\Rightarrow \frac{y^2}{a^2} - \frac{{(-x)}^2}{b^2} =1 \)
   
   \(\Rightarrow \frac{y^2}{a^2} - \frac{x^2}{b^2} =1 \)   ...(2)
   
   Interchanging the Lengths of Transverse and Conjugate Axis in equation (2) above we get
   
   \(\frac{y^2}{b^2} - \frac{x^2}{a^2} =1 \)   ...(3)
   
   The equation (3) above gives the Standard Coordinate Equation for Conjugate Hyperbolas to the Hyperbolas given in equation (1) above which are Hyperbolas having their Transverse Axes Parallel to the \(Y\)-Axis, Center at Origin, Length of Semi-Transverse Axis \(b\) and Length of Semi-Conjugate Axis \(a\).
   
   Similarly, as given in Derivation of Standard and Implicit Coordinate Equation for Axis Aligned Hyperbolas, the Standard Coordinate Equation for Hyperbolas having their Transverse Axes Parallel to the \(Y\)-Axis, Center at Origin, Length of Semi-Transverse Axis \(a\) and Length of Semi-Conjugate Axis \(b\) is given as
   
   \(\frac{y^2}{a^2} - \frac{x^2}{b^2} =1 \)   ...(4)
   
   Rotating the equation (4) above Counter Clockwise by \(90^\circ\) we get
   
   \(\frac{{(y \cos 90^\circ - x \sin 90^\circ)} ^2}{a^2} - \frac{{(x \cos 90^\circ + y \sin 90^\circ)} ^2}{b^2} =1 \)
   
   \(\Rightarrow \frac{{(-x)}^2}{a^2} - \frac{y^2}{b^2} =1 \)
   
   \(\Rightarrow \frac{x^2}{a^2} - \frac{y^2}{b^2} =1 \)   ...(5)
   
   Interchanging the Lengths of Transverse and Conjugate Axis in equation (5) above we get
   
   \(\frac{x^2}{b^2} - \frac{y^2}{a^2} =1 \)   ...(6)
   
   The equation (6) above gives the Standard Coordinate Equation for Conjugate Hyperbolas to the Hyperbolas given in equation (4) above which are Hyperbolas having their Transverse Axes Parallel to the \(X\)-Axis, Center at Origin, Length of Semi-Transverse Axis \(b\) and Length of Semi-Conjugate Axis \(a\).
   
   
3. The following derives the Equation of Rotated Conjugate Hyperbola from Standard Coordinate Equation of Conjugate Hyperbola given in equation (3)
   
   Multiplying equation (3) by \(a^2b^2\) on Both Sides we get
   
   \(- b^2x^2 + a^2y^2 =a^2b^2 \)
   
   \(\Rightarrow -b^2x^2 + a^2y^2 - a^2b^2 = 0 \)   ...(7)
   
   Rotating the equation (7) above Counter Clockwise with angle \(\theta\) with respect to Positive Direction of \(X\) Axis we get
   
   \(-b^2{(x \cos\theta + y \sin\theta)}^2 + a^2{(y \cos\theta - x \sin\theta)}^2 - a^2b^2 = 0 \)
   
   \(\Rightarrow -b^2(x^2 \cos^2\theta + y^2 \sin^2\theta + 2xy\sin\theta\cos\theta )+ a^2(y^2\cos^2\theta + x^2\sin^2\theta - 2xy\sin\theta\cos\theta ) - a^2b^2 = 0 \)
   
   \(\Rightarrow -b^2x^2\cos^2\theta - b^2y^2 \sin^2\theta - 2b^2xy\sin\theta\cos\theta + a^2y^2\cos^2\theta + a^2x^2\sin^2\theta - 2a^2xy\sin\theta\cos\theta - a^2b^2 = 0 \)
   
   \(\Rightarrow (a^2\sin^2\theta - b^2\cos^2\theta )x^2 - (2b^2\sin\theta\cos\theta + 2a^2\sin\theta\cos\theta)xy + (a^2 \cos^2\theta - b^2\sin^2\theta)y^2 - a^2b^2 = 0 \)
   
   \(\Rightarrow (a^2\sin^2\theta - b^2\cos^2\theta )x^2 - (2\sin\theta\cos\theta(b^2 + a^2))xy + (a^2 \cos^2\theta - b^2\sin^2\theta)y^2 - a^2b^2 = 0\)
   
   \(\Rightarrow (a^2\sin^2\theta - b^2\cos^2\theta )x^2 - ((b^2 + a^2)\sin2\theta)xy + (a^2 \cos^2\theta - b^2\sin^2\theta)y^2 - a^2b^2 = 0\)   ...(8)
   
   The equation (8) above gives the Equation of the Rotated Conjugate Hyperbola having Center at Origin
   
   Setting \(a^2\sin^2\theta - b^2\cos^2\theta =\mathbf{A_c}\), \(-(b^2 + a^2)\sin2\theta=\mathbf{B_c}\) and \(a^2\cos^2\theta - b^2 \sin^2\theta=\mathbf{C_c}\) in equation (8) above we get
   
   \(A_cx^2 + B_cxy + C_cy^2 - a^2b^2 = 0\)   ...(9)
   
   Now, Translating the Equation of the Rotated Hyperbola as given in equation (9) above so that it's Center is at \((x_c,y_c)\) we get
   
   \(A_c{(x-x_c)}^2 + B_c(x-x_c)(y-y_c) + C_c{(y-y_c)}^2 - a^2b^2 = 0\)
   
   \(\Rightarrow A_c(x^2+{x_c}^2 - 2x_cx) + B_c(xy-x_cy-y_cx+x_cy_c) + C_c(y^2+{y_c}^2-2y_cy) - a^2b^2 = 0\)
   
   \(\Rightarrow A_cx^2+A_c{x_c}^2 - 2A_cx_cx + B_cxy-B_cx_cy-B_cy_cx+B_cx_cy_c + C_cy^2+C_c{y_c}^2-2C_cy_cy - a^2b^2 = 0\)
   
   \(\Rightarrow A_cx^2+ B_cxy + C_cy^2 - (2A_cx_c+B_cy_c)x -(2C_cy_c+B_cx_c)y +A_c{x_c}^2+C_c{y_c}^2+B_cx_cy_c -a^2b^2=0\)   ...(10)
   
   Setting \(- (2A_cx_c+B_cy_c)=\mathbf{D_c}\), \(-(2C_cy_c+B_cx_c)=\mathbf{E_c}\) and \(A_c{x_c}^2+C_c{y_c}^2+B_cx_cy_c -a^2b^2=\mathbf{F_c}\) in equation (10) above we get
   
   \(A_cx^2 + B_cxy + C_cy^2 +D_cx +E_cy +F_c=0\)   ...(11)
   
   The equation (10) and (11) above give the Implicit Coordinate Equation of the Conjugate Hyperbola for any Hyperbola Rotated Counter Clockwise by Angle \(\theta\) with respect to Positive Direction of \(X\) Axis and having Center at \((x_c,y_c)\) where
   
   \(A_c = a^2\sin^2\theta - b^2\cos^2\theta\)
   
   \(B_c = -(b^2 + a^2)\sin2\theta\)
   
   \(C_c = a^2\cos^2\theta - b^2 \sin^2\theta\)
   
   \(D_c = -(2A_cx_c+B_cy_c) = -(2(a^2\sin^2\theta - b^2\cos^2\theta) x_c - ((b^2 + a^2)\sin2\theta) y_c) = 2(b^2\cos^2\theta - a^2\sin^2\theta) x_c + ((b^2 + a^2)\sin2\theta) y_c\)
   
   \(E_c = -(2C_cy_c+B_cx_c) = -(2(a^2\cos^2\theta - b^2\sin^2\theta) y_c - ((b^2 + a^2)\sin2\theta) x_c) = 2(b^2\sin^2\theta - a^2\cos^2\theta) y_c + ((b^2 + a^2)\sin2\theta) x_c\)
   
   \(F_c = A_c{x_c}^2+C_c{y_c}^2+B_cx_cy_c -a^2b^2 = (a^2\sin^2\theta - b^2\cos^2\theta) {x_c}^2 + (a^2\cos^2\theta - b^2\sin^2\theta) {y_c}^2 - ((b^2 + a^2)\sin2\theta) x_cy_c - a^2b^2\)   ...(12)
   
   Now, as given in Derivation of Implicit Coordinate Equation for Arbitrarily Rotated and Translated Hyperbolas, the Implicit Coordinate Equation of any for Arbitrarily Rotated and Translated Hyperbola is given as
   
   \(Ax^2 + Bxy + Cy^2 +Dx +Ey +F=0\)   ...(13)
   
   where
   
   \(A = b^2\cos^2\theta - a^2\sin^2\theta\)
   
   \(B = (b^2 + a^2)\sin2\theta\)
   
   \(C = b^2 \sin^2\theta - a^2\cos^2\theta\)
   
   \(D = -(2Ax_c+By_c) = -(2(b^2\cos^2\theta - a^2\sin^2\theta) x_c + ((b^2 + a^2)\sin2\theta) y_c)\)
   
   \(E = -(2Cy_c+Bx_c) = -(2(b^2 \sin^2\theta - a^2\cos^2\theta) y_c + ((b^2 + a^2)\sin2\theta) x_c)\)
   
   \(F = A{x_c}^2+C{y_c}^2+Bx_cy_c -a^2b^2 = (b^2\cos^2\theta - a^2\sin^2\theta) {x_c}^2 + (b^2 \sin^2\theta - a^2\cos^2\theta) {y_c}^2 + ((b^2 + a^2)\sin2\theta) x_cy_c - a^2b^2\)   ...(14)
   
   From equation sets (12) and (14) we get
   
   \(A_c= -A,\hspace{3mm}B_c= -B,\hspace{3mm}C_c= -C,\hspace{3mm}D_c= -D,\hspace{3mm}E_c= -E,\hspace{3mm}F_c= -A{x_c}^2-C{y_c}^2-Bx_cy_c -a^2b^2\)   ...(15)
   
   Now, \(F_c\) can be also be given as
   
   \(F_c = F + P\)   ...(16)
   
   where \(P\) is a constant
   
   \(\Rightarrow P = F_c-F\)   ...(17)
   
   Substituting the value of \(F\) and \(F_c\) from equations sets (14) and (15) respectively in equation (17) above we get
   
   \(P = -A{x_c}^2-C{y_c}^2-Bx_cy_c -a^2b^2 - A{x_c}^2-C{y_c}^2-Bx_cy_c + a^2b^2\)
   
   \(\Rightarrow P = -2A{x_c}^2-2C{y_c}^2-2Bx_cy_c\)
   
   \(\Rightarrow P = -2A{x_c}^2-Bx_cy_c -2C{y_c}^2-Bx_cy_c\)
   
   \(\Rightarrow P = -(2A{x_c}+By_c)x_c -(2Cy_c+Bx_c)y_c\)   ...(18)
   
   Since, as given in equation set (14) \(D= -(2A{x_c}+By_c)\) and \(E=-(2C{y_c}+Bx_c)\). Substituting this in equation (18) above we get
   
   \(P = Dx_c + Ey_c\)   ...(19)
   
   Now, substituting the value of \(P\) from equation (19) in equation (16) we get
   
   \(F_c = F + Dx_c + Ey_c\)   ...(20)
   
   Hence, Using equation set (15) and equation (20) above, for any Arbitrarily Rotated and Translated Hyperbola whose Implicit Equation is given by equation (10) above, the Implicit Equation of it's corresponding Conjugate Hyperbola is given as
   
   \(-Ax^2 - Bxy - Cy^2 - Dx - Ey + F + Dx_c + Ey_c=0\)   ...(21)