Question

Chords of a hyperbola are drawn, all passing through the fixed point $(h, k)$ prove that the locus of their middle points is a hyperbola whose centre is the point $(\frac{h}{2}, \frac{k}{2})$ and which is similar to either the hyperbola or its conjugate.

Answer 1

Locus is a hyperbola with center $(\frac{h}{2}, \frac{k}{2})$, similar to the original hyperbola or its conjugate

Answer 2

Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Let the fixed point through which the chords pass be $P(h, k)$. Let $M(x_1, y_1)$ be the middle point of a chord of the hyperbola. The equation of the chord of the hyperbola with middle point $M(x_1, y_1)$ is given by $T = S_1$. Here, $T = \frac{xx_1}{a^2} - \frac{yy_1}{b^2} - 1$ and $S_1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1$. The equation of the chord is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} - 1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1$, which simplifies to $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}$.

Since the chord passes through the fixed point $(h, k)$, the coordinates $(h, k)$ must satisfy the equation of the chord: $\frac{hx_1}{a^2} - \frac{ky_1}{b^2} = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}$.

The locus of the middle point $(x_1, y_1)$ is obtained by replacing $(x_1, y_1)$ with $(x, y)$: $\frac{hx}{a^2} - \frac{ky}{b^2} = \frac{x^2}{a^2} - \frac{y^2}{b^2}$. Rearranging the terms, we get: $\frac{x^2}{a^2} - \frac{hx}{a^2} - \frac{y^2}{b^2} + \frac{ky}{b^2} = 0$.

Completing the square for the $x$ and $y$ terms: $\frac{1}{a^2}(x^2 - hx) - \frac{1}{b^2}(y^2 - ky) = 0$. $\frac{1}{a^2}((x - \frac{h}{2})^2 - (\frac{h}{2})^2) - \frac{1}{b^2}((y - \frac{k}{2})^2 - (\frac{k}{2})^2) = 0$. $\frac{(x - \frac{h}{2})^2}{a^2} - \frac{h^2}{4a^2} - \frac{(y - \frac{k}{2})^2}{b^2} + \frac{k^2}{4b^2} = 0$. $\frac{(x - \frac{h}{2})^2}{a^2} - \frac{(y - \frac{k}{2})^2}{b^2} = \frac{h^2}{4a^2} - \frac{k^2}{4b^2}$.

Let $X = x - \frac{h}{2}$ and $Y = y - \frac{k}{2}$. Let $C = \frac{h^2}{4a^2} - \frac{k^2}{4b^2}$. The equation of the locus is $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = C$.

This is the equation of a conic section with center at $X=0, Y=0$, which corresponds to $x = \frac{h}{2}$ and $y = \frac{k}{2}$. Thus, the center of the locus is $(\frac{h}{2}, \frac{k}{2})$.

The nature of the conic depends on the value of $C$.

Case 1: $C > 0$. The equation is $\frac{X^2}{a^2 C} - \frac{Y^2}{b^2 C} = 1$. This is a hyperbola with semi-axes $A = a\sqrt{C}$ and $B = b\sqrt{C}$. The ratio of the squares of the semi-axes is $\frac{B^2}{A^2} = \frac{b^2 C}{a^2 C} = \frac{b^2}{a^2}$. The original hyperbola has the ratio of squares of semi-axes $\frac{b^2}{a^2}$. Since the ratio is the same, the locus hyperbola is similar to the original hyperbola.

Case 2: $C < 0$. Let $C' = -C > 0$. The equation is $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = -C'$, or $\frac{Y^2}{b^2} - \frac{X^2}{a^2} = C'$. Dividing by $C'$, we get $\frac{Y^2}{b^2 C'} - \frac{X^2}{a^2 C'} = 1$. This is a hyperbola with the transverse axis along the Y-axis. Its semi-axes are $B' = b\sqrt{C'}$ and $A' = a\sqrt{C'}$. The ratio of the squares of the semi-axes is $\frac{A'^2}{B'^2} = \frac{a^2 C'}{b^2 C'} = \frac{a^2}{b^2}$. The conjugate hyperbola to the original hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, which has the ratio of squares of semi-axes $\frac{a^2}{b^2}$. Since the ratio is the same, the locus hyperbola is similar to the conjugate hyperbola.

Case 3: $C = 0$. The equation is $\frac{(x - \frac{h}{2})^2}{a^2} - \frac{(y - \frac{k}{2})^2}{b^2} = 0$, which represents a pair of straight lines $\frac{x - \frac{h}{2}}{a} \pm \frac{y - \frac{k}{2}}{b} = 0$ passing through the center $(\frac{h}{2}, \frac{k}{2})$. These lines are the asymptotes of the hyperbola $\frac{(x - \frac{h}{2})^2}{a^2} - \frac{(y - \frac{k}{2})^2}{b^2} = C$. The slopes of these lines are $\pm \frac{b}{a}$, which are the same as the slopes of the asymptotes of the original hyperbola. A pair of intersecting lines can be considered a degenerate hyperbola, and its shape is determined by the angle between the lines, which is related to the ratio of the semi-axes of the non-degenerate hyperbola. Since the slopes are the same, the angle between the lines is the same as the angle between the asymptotes of the original hyperbola, implying similarity in shape.

Thus, the locus of the middle points is a hyperbola whose center is $(\frac{h}{2}, \frac{k}{2})$ and which is similar to either the hyperbola or its conjugate.