Question

Prove that the product of the distance from any point on a hyperbola to its
asymptotes is a constant.

Answer 1

Answer on Question #78658 – Math – Analytic Geometry Question

Prove that the product of the distance from any point on a hyperbola to its asymptotes is a constant.

Solution

Let there be given a hyperbola. If the axes of a rectangular coordinate system are chosen so that the foci of the given hyperbola are symmetrically situated on the $x$-axis with respect to the origin, then the equation of the hyperbola has the form

$$\frac {x ^ {2}}{a ^ {2}} - \frac {y ^ {2}}{b ^ {2}} = 1$$

The equations of the asymptotes are

$$y = \frac {b}{a} x, \quad y = - \frac {b}{a} x$$

or

$$t _ {1} \colon b x - a y = 0, \quad t _ {2} \colon b x + a y = 0$$

Let $P(x_0, y_0)$ be any point on a hyperbola. Then the distance $d_1$ from $P$ to $t_1$ is given by

$$d _ {1} = \frac {\left| b x _ {0} - a y _ {0} \right|}{\sqrt {a ^ {2} + b ^ {2}}}$$

The distance $d_{2}$ from $\mathbf{P}$ to $t_2$ is given by

$$d _ {2} = \frac {\left| b x _ {0} + a y _ {0} \right|}{\sqrt {a ^ {2} + b ^ {2}}}$$

Hence

$$d _ {1} d _ {2} = \frac {\left| b x _ {0} - a y _ {0} \right|}{\sqrt {a ^ {2} + b ^ {2}}} \cdot \frac {\left| b x _ {0} + a y _ {0} \right|}{\sqrt {a ^ {2} + b ^ {2}}} = \frac {\left| b ^ {2} x _ {0} ^ {2} - a ^ {2} y _ {0} ^ {2} \right|}{a ^ {2} + b ^ {2}}$$

But $P(x_0, y_0)$ is the point on the hyperbola. Then

$$\frac {x _ {0} ^ {2}}{a ^ {2}} - \frac {y _ {0} ^ {2}}{b ^ {2}} = 1 \Rightarrow b ^ {2} x _ {0} ^ {2} - a ^ {2} y _ {0} ^ {2} = a ^ {2} b ^ {2}$$

Therefore

$$d _ {1} d _ {2} = \frac {\left| b ^ {2} x _ {0} ^ {2} - a ^ {2} y _ {0} ^ {2} \right|}{a ^ {2} + b ^ {2}} = \frac {a ^ {2} b ^ {2}}{a ^ {2} + b ^ {2}} = \text{const}$$

We prove that the product of the distance from any point on a hyperbola to its asymptotes is a constant.

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