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Let's say we have a hyperbola with the focus at $F=(2, 3)$, directrix of $y=-x$, and eccentricity $e=\sqrt{2}$. If I understand correctly, that should mean that the distance from any point $P$ on the hyperbola to the focus is $\sqrt{2}$ times larger than the distance between $P$ and the directrix.

Since point $(0, 0)$ is on the directrix, and the directrix normal vector is $(\frac{1}{\sqrt2}, \frac{1}{\sqrt2})$, the hyperbola can be represented as $\|(x-2)^2 + (y-3)^2\| = \sqrt{2} |\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}|=|x+y|$.

Let's take a point on hyperbola $P = (1, \frac{11}{4})$.

The distance from that point to the focus is $d_1 = \sqrt{(1-2)^2 + (\frac{11}{4} - 3)^2} = \sqrt{1 + \frac{1}{16}} = \frac{\sqrt{17}}{4}$.

The (shortest) distance between the point and the directrix is $d_2 = |(1, \frac{11}{4})(\frac{1}{\sqrt2}, \frac{1}{\sqrt{2}})| = \frac{15}{4\sqrt2}$.

Why is $d_1 \ne \sqrt{2} d_2$? Where did I make a mistake this time?

S11n
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1 Answers1

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The equation of hyperbola is $2xy+4x+6y=13$

Mistake - The point chosen by you $(1,\frac{11}{4})$ does not lie on the hyperbola.

To verify your claim " the distance from any point $P$ on the hyperbola to the focus is $\sqrt{2}$ times larger than the distance between $P$ and the directrix. "

Let us take a point P as P$(2,\frac{1}{2})$.

The distance of P from the directrix $x+y=0$ is $\frac{5\sqrt2}{4}$----(1)

The distance of P from focus $(2,3)$ is $\frac{5}{2}$----(2)

Eccentricity = $\sqrt 2$----(3)

As (2)=(1)(3),

It is verified $d_1=\sqrt2d_2$

NadiKeUssPar
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