Derivation of Standard and Implicit Coordinate Equation for Axis Aligned Hyperbolas

The following gives the derivation for Standard and Implicit Coordinate Equations for any Hyperbola having it's Transverse Axis Parallel to the \(X\)-Axis.

Let's consider a Hyperbola having it's Semi-Transverse Axis Length as \(a\), Semi-Conjugate Axis Length as \(b\), Coordinates of Center at \((x_c,y_c)\), and Coordinates of 2 Foci at \((x_{f1},y_f)\) and \((x_{f2},y_f)\). Now, 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 \(2a\). Hence

\(\sqrt{{(x-x_{f1})}^2 + {(y-y_f)}^2} - \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2} =L=2a\)

\(\Rightarrow \sqrt{{(x-x_{f1})}^2 + {(y-y_f)}^2} = 2a + \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)   ...(1)

Squaring Both Sides of equation (1) above we get

\({(x-x_{f1})}^2 + {(y-y_f)}^2 = 4a^2 + {(x-x_{f2})}^2 + {(y-y_f)}^2 + 4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow 4a^2 + {(x-x_{f2})}^2 - {(x-x_{f1})}^2 = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow 4a^2 + x^2 + {x_{f2}}^2 - 2x_{f2}x - x^2 - {x_{f1}}^2 + 2x_{f1}x = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow 4a^2 + {x_{f2}}^2 - {x_{f1}}^2 - 2x_{f2}x + 2x_{f1}x = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow 4a^2 + ({x_{f2} - x_{f1}})({x_{f2} + x_{f1}}) - 2x(x_{f2}-x_{f1}) = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)   ...(2)

Now, \({x_{f2} - x_{f1}}=F=2c\) (i.e Distance Between the 2 Foci) and \({x_{f2} + x_{f1}}=2x_c\) (i.e 2 times the \(X\)-Coordinate of the Center of Hyperbola). Substituting these in equation (2) above we get

\(\Rightarrow 4a^2 + (2c)(2x_c) - 2x(2c) = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow 4a^2 + 4cx_c - 4cx = -4a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)

\(\Rightarrow a^2 + cx_c - cx = -a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2}\)   ...(3)

Squaring Both Sides of equation (3) above we get

\({(a^2 + cx_c - cx)}^2 = {(-a \sqrt{{(x-x_{f2})}^2 + {(y-y_f)}^2})}^2\)

\(\Rightarrow a^4 + c^2{x_c}^2 + c^2x^2 + 2a^2cx_c - 2a^2cx - 2c^2x_cx = a^2{(x-x_{f2})}^2 + a^2{(y-y_f)}^2\)

\(\Rightarrow c^2x^2 - 2a^2cx - 2c^2x_cx + a^4 + c^2{x_c}^2 + 2a^2cx_c = a^2x^2 + a^2{x_{f2}}^2 - 2a^2x_{f2}x + a^2{(y-y_f)}^2\)

\(\Rightarrow c^2x^2 - a^2x^2 + 2a^2x_{f2}x - 2a^2cx - 2c^2x_cx + a^4 + c^2{x_c}^2 + 2a^2cx_c - a^2{x_{f2}}^2 = a^2{(y-y_f)}^2\)

\(\Rightarrow (c^2 - a^2)x^2 + (2a^2x_{f2} - 2a^2c - 2c^2x_c)x + a^4 + c^2{x_c}^2 + 2a^2cx_c - a^2{x_{f2}}^2 = a^2{(y-y_f)}^2\)   ...(4)

Now, for any Hyperbola we know that \(a^2 + b^2 = c^2\hspace{3mm}\Rightarrow b^2=c^2-a^2\). Also, since the Transverse Axis of Hyperbola is Parallel to the \(X\)-Axis therefore \(y_f=y_c\). Substituting these in equation (4) above we get

\(\Rightarrow b^2x^2 + (2a^2x_{f2} - a^2(x_{f2}-x_{f1}) - 2(a^2+b^2)x_c)x + a^4 + (a^2+b^2){x_c}^2 + 2a^2cx_c - a^2{x_{f2}}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 + (2a^2x_{f2} - a^2x_{f2} + a^2x_{f1} - 2a^2x_c - 2b^2x_c)x + a^4 + a^2{x_c}^2 + 2a^2cx_c - a^2{x_{f2}}^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 + (a^2(x_{f2}+ x_{f1}) - 2a^2x_c - 2b^2x_c)x + a^4 + a^2{(\frac{x_{f2} + x_{f1}}{2})}^2 + \frac{a^2}{2}(x_{f2}- x_{f1})(x_{f2}+ x_{f1}) - a^2{x_{f2}}^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 + (2a^2x_c - 2a^2x_c - 2b^2x_c)x + a^4 + a^2(\frac{{x_{f1}}^2 + {x_{f2}}^2 + 2x_{f1}x_{f2}}{4}) + \frac{a^2}{2}({x_{f2}}^2- {x_{f1}}^2) - a^2{x_{f2}}^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 + a^2(\frac{{x_{f1}}^2 + {x_{f2}}^2 + 2x_{f1}x_{f2}}{4}) + a^2 (\frac{{x_{f2}}^2- {x_{f1}}^2 - 2{x_{f2}}^2}{2}) +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 + a^2(\frac{{x_{f1}}^2 + {x_{f2}}^2 + 2x_{f1}x_{f2}}{4} + \frac{-{x_{f2}}^2- {x_{f1}}^2}{2}) +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 + a^2(\frac{{x_{f1}}^2 + {x_{f2}}^2 + 2x_{f1}x_{f2} -2{x_{f2}}^2 -2{x_{f1}}^2}{4}) +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 - a^2(\frac{{x_{f2}}^2 + {x_{f1}}^2 - 2x_{f1}x_{f2}}{4}) +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 - a^2{(\frac{x_{f2}- x_{f1}}{2})}^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^4 - a^2c^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx + a^2(a^2-c^2) +b^2{x_c}^2 = a^2{(y-y_c)}^2\)

\(\Rightarrow b^2x^2 - 2b^2x_cx - a^2b^2 +b^2{x_c}^2 = a^2{(y-y_c)}^2\)   ...(5)

Now, Dividing By \(a^2b^2\) on Both Sides of equation (5) above we get

\(\frac{x^2}{a^2} - \frac{2x_cx}{a^2} + \frac{{x_c}^2}{a^2} - 1 = \frac{{(y-y_c)}^2}{b^2}\)

\(\Rightarrow \frac{{(x-x_c)}^2}{a^2} - \frac{{(y-y_c)}^2}{b^2} =1 \)   ...(6)

The equation (6) above gives the Standard Coordinate Equation for any Hyperbola having it's Transverse Axis Parallel to the \(X\)-Axis.

For any Hyperbola having its Center at the Origin equation (6) becomes

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} =1 \)   ...(7)

Multiplying By \(a^2b^2\) on Both Sides of equation (6) above we get

\(b^2{(x-x_c)}^2- a^2{(y-y_c)}^2=a^2b^2 \)

\(\Rightarrow b^2(x^2-2x_cx+{x_c}^2)- a^2(y^2-2y_cy+{y_c}^2)=a^2b^2 \)

\(\Rightarrow b^2x^2-2b^2x_cx+b^2{x_c}^2- a^2y^2+2a^2y_cy-a^2{y_c}^2=a^2b^2 \)

\(\Rightarrow b^2x^2- a^2y^2-2b^2x_cx+2a^2y_cy+b^2{x_c}^2-a^2{y_c}^2-a^2b^2=0 \)   ...(8)

Setting \(b^2=\mathbf{A}\), \(-a^2=\mathbf{C}\) in equation (8) above we get

\(Ax^2+ Cy^2- 2Ax_cx - 2Cy_cy +A{x_c}^2+C{y_c}^2 + AC=0 \)   ...(9)

Setting \(-2Ax_c=\mathbf{D}\), \(-2Cy_c=\mathbf{E}\) and \(A{x_c}^2+C{y_c}^2 + AC=\mathbf{F}\) in equation (9) above we get

\(Ax^2+ Cy^2 + Dx + Ey + F=0 \)   ...(10)

The equations (8), (9) and (10) above give the Implicit Coordinate Equation of any Hyperbola having it's Transverse Axis Parallel to the \(X\)-Axis.

For any Hyperbola having its Center at the Origin equation (10) becomes

\(Ax^2+ Cy^2+F=0 \)   ...(11)
The following gives the derivation for Standard and Implicit Coordinate Equations for any Hyperbola having it's Transverse Axis Parallel to the \(Y\)-Axis.

Let's consider a Hyperbola having it's Semi-Transverse Axis Length as \(a\), Semi-Conjugate Axis Length as \(b\), Coordinates of Center at \((x_c,y_c)\), and Coordinates of 2 Foci at \((x_{f1},y_f)\) and \((x_{f2},y_f)\). Now, 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 \(2a\). Hence

\(\sqrt{{(x-x_f)}^2 + {(y-y_{f1})}^2} - \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2} =L=2a\)

\(\Rightarrow \sqrt{{(x-x_f)}^2 + {(y-y_{f1})}^2} = 2a + \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)   ...(12)

Squaring Both Sides of equation (12) above we get

\({(x-x_f)}^2 + {(y-y_{f1})}^2 = 4a^2 + {(x-x_f)}^2 + {(y-y_{f2})}^2 + 4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow 4a^2 + {(y-y_{f2})}^2 - {(y-y_{f1})}^2 = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow 4a^2 + y^2 + {y_{f2}}^2 - 2y_{f2}y - y^2 - {y_{f1}}^2 + 2y_{f1}y = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow 4a^2 + {y_{f2}}^2 - {y_{f1}}^2 - 2y_{f2}y + 2y_{f1}y = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow 4a^2 + ({y_{f2} - y_{f1}})({y_{f2} + y_{f1}}) - 2y(y_{f2}-y_{f1}) = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)   ...(13)

Now, \({y_{f2} - y_{f1}}=F=2c\) (i.e Distance Between the 2 Foci) and \({y_{f2} + y_{f1}}=2y_c\) (i.e 2 times the \(Y\)-Coordinate of the Center of Hyperbola). Substituting these in equation (13) above we get

\(\Rightarrow 4a^2 + (2c)(2y_c) - 2y(2c) = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow 4a^2 + 4cy_c - 4cy = -4a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)

\(\Rightarrow a^2 + cy_c - cy = -a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2}\)   ...(14)

Squaring Both Sides of equation (14) above we get

\({(a^2 + cy_c - cy)}^2 = {(-a \sqrt{{(x-x_f)}^2 + {(y-y_{f2})}^2})}^2\)

\(\Rightarrow a^4 + c^2{y_c}^2 + c^2y^2 + 2a^2cy_c - 2a^2cy - 2c^2y_cy = a^2{(y-y_{f2})}^2 + a^2{(x-x_f)}^2\)

\(\Rightarrow c^2y^2 - 2a^2cy - 2c^2y_cy + a^4 + c^2{y_c}^2 + 2a^2cy_c = a^2y^2 + a^2{y_{f2}}^2 - 2a^2y_{f2}y + a^2{(x-x_f)}^2\)

\(\Rightarrow c^2y^2 - a^2y^2 + 2a^2y_{f2}x - 2a^2cy - 2c^2y_cy + a^4 + c^2{y_c}^2 + 2a^2cy_c - a^2{y_{f2}}^2 = a^2{(x-x_f)}^2\)

\(\Rightarrow (c^2 - a^2)y^2 + (2a^2x_{f2} - 2a^2c - 2c^2x_c)y + a^4 + c^2{y_c}^2 + 2a^2cy_c - a^2{y_{f2}}^2 = a^2{(x-x_f)}^2\)   ...(15)

Now, for any Hyperbola we know that \(a^2 + b^2 = c^2\hspace{3mm}\Rightarrow b^2=c^2-a^2\). Also, since the Transverse Axis of Hyperbola is Parallel to the \(Y\)-Axis therefore \(x_f=x_c\). Substituting these in equation (15) above we get

\(\Rightarrow b^2y^2 + (2a^2y_{f2} - a^2(y_{f2}-y_{f1}) - 2(a^2+b^2)y_c)y + a^4 + (a^2+b^2){y_c}^2 + 2a^2cy_c - a^2{y_{f2}}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 + (2a^2y_{f2} - a^2y_{f2} + a^2y_{f1} - 2a^2y_c - 2b^2y_c)x + a^4 + a^2{y_c}^2 + 2a^2cy_c - a^2{y_{f2}}^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 + (a^2(y_{f2}+ y_{f1}) - 2a^2y_c - 2b^2y_c)x + a^4 + a^2{(\frac{y_{f2} + y_{f1}}{2})}^2 + \frac{a^2}{2}(y_{f2}- y_{f1})(y_{f2}+ y_{f1}) - a^2{y_{f2}}^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 + (2a^2y_c - 2a^2y_c - 2b^2y_c)y + a^4 + a^2(\frac{{y_{f1}}^2 + {y_{f2}}^2 + 2y_{f1}y_{f2}}{4}) + \frac{a^2}{2}({y_{f2}}^2- {y_{f1}}^2) - a^2{y_{f2}}^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 + a^2(\frac{{y_{f1}}^2 + {y_{f2}}^2 + 2y_{f1}y_{f2}}{4}) + a^2 (\frac{{y_{f2}}^2- {y_{f1}}^2 - 2{y_{f2}}^2}{2}) +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 + a^2(\frac{{y_{f1}}^2 + {y_{f2}}^2 + 2y_{f1}y_{f2}}{4} + \frac{-{y_{f2}}^2- {y_{f1}}^2}{2}) +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 + a^2(\frac{{y_{f1}}^2 + {y_{f2}}^2 + 2y_{f1}y_{f2} -2{y_{f2}}^2 -2{y_{f1}}^2}{4}) +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 - a^2(\frac{{y_{f2}}^2 + {y_{f1}}^2 - 2y_{f1}y_{f2}}{4}) +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 - a^2{(\frac{y_{f2}- y_{f1}}{2})}^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^4 - a^2c^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy + a^2(a^2-c^2) +b^2{y_c}^2 = a^2{(x-x_c)}^2\)

\(\Rightarrow b^2y^2 - 2b^2y_cy - a^2b^2 +b^2{y_c}^2 = a^2{(x-x_c)}^2\)   ...(16)

Now, Dividing By \(a^2b^2\) on Both Sides of equation (16) above we get

\(\frac{y^2}{a^2} - \frac{2y_cy}{a^2} + \frac{{y_c}^2}{a^2} - 1 = \frac{{(x-x_c)}^2}{b^2}\)

\(\Rightarrow \frac{{(y-y_c)}^2}{a^2} - \frac{{(x-x_c)}^2}{b^2} =1 \)   ...(17)

The equation (17) above gives the Standard Coordinate Equation for any Hyperbola having it's Transverse Axis Parallel to the \(Y\)-Axis.

For any Hyperbola having its Center at the Origin equation (17) becomes

\(\frac{y^2}{a^2} - \frac{x^2}{b^2} =1 \)   ...(18)

Multiplying By \(a^2b^2\) on Both Sides of equation (17) above we get

\(b^2{(y-y_c)}^2 - a^2{(x-x_c)}^2 =a^2b^2 \)

\(\Rightarrow b^2(y^2-2y_cy+{y_c}^2) - a^2(x^2-2x_cx+{x_c}^2) =a^2b^2 \)

\(\Rightarrow -a^2x^2+2a^2x_cx-a^2{x_c}^2+ b^2y^2-2b^2y_cy+b^2{y_c}^2=a^2b^2 \)

\(\Rightarrow -a^2x^2+ b^2y^2+2a^2x_cx-2b^2y_cy-a^2{x_c}^2+b^2{y_c}^2-a^2b^2=0 \)   ...(19)

Setting \(-a^2=\mathbf{A}\), \(b^2=\mathbf{C}\) in equation (18) above we get

\(Ax^2+ Cy^2- 2Ax_cx - 2Cy_cy +A{x_c}^2+C{y_c}^2+ AC=0 \)   ...(20)

Setting \(-2Ax_c=\mathbf{D}\), \(-2Cy_c=\mathbf{E}\) and \(A{x_c}^2+C{y_c}^2+AC=\mathbf{F}\) in equation (20) above we get

\(Ax^2+ Cy^2 + Dx + Ey + F=0 \)   ...(21)

The equations (19), (20) and (21) above give the Implicit Coordinate Equation of any Hyperbola having it's Transverse Axis Parallel to the \(Y\)-Axis.

For any Hyperbola having its Center at the Origin equation (21) becomes

\(Ax^2+ Cy^2+F=0 \)   ...(22)
Following are the Properties of Implicit Equation of Axis Aligned Hyperbolas
1. The Sign of Co-efficients \(A\) and \(C\) are always different.
2. In equations (11) and (22) the Axis having Co-efficient that has the Same Sign as the Constant \(F\) forms the Conjugate-Axis and the Axis having Co-efficient that has a Different Sign than that of Constant \(F\) forms the Transverse-Axis. If the Co-efficient \(A\) has Different Sign than that of the Constant \(F\), then the Equations represent a \(X\)-Transverse Hyperbola. Similarly, if the Co-efficient \(C\) has Different Sign than of the Constant \(F\), then the Equations represents a \(Y\)-Transverse Hyperbola.

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