Academy
Formulas/maths/Hyperbola/Chord with a Given Midpoint

Chord with a Given Midpoint

Chord of x²/a² − y²/b² = 1 whose midpoint is (x₁, y₁). Explicitly: xx₁/a² − yy₁/b² − 1 = x₁²/a² − y₁²/b² − 1.
Derivation

Let (x1,y1)(x_1, y_1) be the midpoint of a chord of x2/a2y2/b2=1x^2/a^2 - y^2/b^2 = 1 with endpoints (x2,y2)(x_2, y_2) and (x3,y3)(x_3, y_3).

Both on the hyperbola:

x22x32a2y22y32b2=0\frac{x_2^2-x_3^2}{a^2} - \frac{y_2^2-y_3^2}{b^2} = 0 (x2+x3)(x2x3)a2=(y2+y3)(y2y3)b2\frac{(x_2+x_3)(x_2-x_3)}{a^2} = \frac{(y_2+y_3)(y_2-y_3)}{b^2}

Using midpoint: x2+x3=2x1x_2+x_3 = 2x_1, y2+y3=2y1y_2+y_3 = 2y_1:

Slope of chord=y2y3x2x3=b2x1a2y1\text{Slope of chord} = \frac{y_2-y_3}{x_2-x_3} = \frac{b^2x_1}{a^2y_1}

(Note: positive slope for the hyperbola, compared to negative for the ellipse — the minus sign in the hyperbola equation reverses the sign.)

Chord through (x1,y1)(x_1,y_1) with this slope:

yy1=b2x1a2y1(xx1)y - y_1 = \frac{b^2x_1}{a^2y_1}(x-x_1)

Rearranging: T=S1T = S_1, where T=xx1/a2yy1/b21T = xx_1/a^2 - yy_1/b^2 - 1 and S1=x12/a2y12/b21S_1 = x_1^2/a^2 - y_1^2/b^2 - 1.