Interesting association between tangent lines of slope one and ellipses Why is it that a tangent line with slope $1$ to an ellipse centered at the origin will have a transformation of $\pm \sqrt{a^2 +b^2}$ where $a$ and $b$ are the major and minor axis of the ellipse?
For example: The tangent line of slope one to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is $y=x+\sqrt{13}$.

The relation you are looking for can be derived algebraically. Start with some general line $y=mx+c$ for the tangent. Substitute it into the equation of your ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This gives:
$$
\frac{x^2}{a^2} + \frac{(mx+c)^2}{b^2} = 1 \\
b^2 x^2 + a^2 (m^2 x^2 + 2mxc + c^2) = a^2 b^2 \\
(b^2 + m^2 a^2)x^2 + 2a^2 mc x + a^2 (c^2 - b^2) = 0 \\
$$
Now since the line is a tangent the discriminant of this quadratic equation will be zero:
$$
\Delta = (2a^2 mc)^2 - 4(b^2 + a^2 m^2)[a^2(c^2 - b^2)] = 0 \\
$$
Rearrange it to obtain the value of $c$:
$$
4a^4 m^2 c^2 - 4a^2 (b^2 c^2 - b^4 + a^2 m^2 c^2 - a^2 m^2 b^2 ) = 0 \\
b^2 c^2 - b^4 - a^2 m^2 b^2 = 0 \\
b^2 c^2 = b^4 + a^2 m^2 b^2 \\
c^2 = b^2 + a^2 m^2 \\
\therefore c = \pm \sqrt{a^2 m^2 + b^2}
$$
So the tangent line equation is $y = mx \pm \sqrt{a^2 m^2 + b^2}$. In your case when the gradient $m=1$ then we have the required transformation $y=x \pm \sqrt{a^2 + b^2}$. 
NB: Geometrically speaking the tangents to the ellipse pass through the foci of a hyperbola with the same $a$ and $b$ values e.g. for the case of $a=3, b=2$.

