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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.
Related Calculators
Conic from Foci and Major/Transverse Axis Length Calculator
Related Topics
Finding Equation of Hyperbola from a given Focus, a Vertex and Eccentricity,    Finding Equation of Hyperbola from given Adjacent Focus, Directrix and Eccentricity,    Finding Parametric Equations for Axis Aligned and Rotated Hyperbola Based on Secant and Tangent Ratios,    Finding Parametric Equations for Axis Aligned and Rotated Hyperbola Based on Hypebolic Sine and Cosine,    Introduction to Hyperbola,    General Quadratic Equations in 2 Variables and Conic Sections
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