# hyperbola eccentricity in a concrete example not working out

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?

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 $$d_1=\sqrt2d_2$$