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 $y=x+\sqrt{13}$.
 A: 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 $a=3, b=2$.

A: This has association with the director circle. This is well known that locus of intersection of perpendicular tangents is $$x^2+y^2=\sqrt{a^2+b^2}$$

Now draw a tangent of slope $m$ in first quadrant of ellipse. Now draw a tangent at lower symmetrical point in fourth quadrant.

As you can see that they are mirror images of each other, mirror being the x-axis. Their slopes are additive inverses, i.e., negative of each other.
You want one slope to be $1$, the other $-1$ you get for no charge!
It is another very well known fact that two lines are perpendicular if $$m_1m_2=-1$$
Clearly, $1$ and $-1$ are possible solutions!
This will look a bit like :

So, that is how you end up with that mysterious factor of $\sqrt{a^2+b^2}$ every time you take $m=1$. The algebraic $x$ and $y$ intercepts come out to be $\pm\sqrt{a^2+b^2}$.
A: It's actually quite simple.  An ellipse centered at the origin with semimajor and semiminor axes $a$ and $b$ can be transformed to a circle by a suitable scaling of the horizontal axis by a factor of $b/a$.  Since such a linear transformation preserves tangency, the resulting line has a slope that is also scaled by the same factor; i.e., if $y = x + \beta$ is the tangent line, then the transformed line after the scaling has the form $y' = \frac{a}{b}x' + \beta$, where $(x',y') = (\frac{b}{a}x, \frac{b}{a}y)$.  Now, such a line is tangent to the circle $(x')^2 + (y')^2 = b^2$, so we have by similarity of right triangles $$\frac{a}{b} = \frac{\sqrt{\beta^2 - b^2}}{b},$$ or equivalently, $$\beta^2 = a^2 + b^2,$$ and we are done.
A: I am going to solve the slightly modified (but essentially equivalent) problem of finding the tangent line with slope $-1$.
We can parameterize the ellipse as $$x(t)=(a\cos t,b\sin t)$$ Then we are interested in the point where its derivative will be pointing perpendicular to the vector $(1,1)$ so that the slope of the tangent there is $-1$. $$\begin{align}x'(t)\cdot (1,1)=0 \\ (-a\sin t,b\cos t)\cdot (1,1)=0 \\ a\sin t-b\cos t=0\end{align}$$ This implies that the point of tangency will occur when $$\tan t={b\over a}$$

In the figure above I have constructed triangle $\Delta HFG$ with legs of length $a$ and $b$, so the angle $\angle GHF=t$. 
If we draw a circle centered at the origin through the endpoint of the semi-minor axis of the ellipse, then it will have radius $b$ (where $b<a$). It will therefore intersect the line $HG$ at the point $J=b(\cos t,\sin t)$. 

Note that the y-coordinate of this point is identical to the y-coordinate of the point $x(t)$ on the ellipse. Thus we can find $x(t)$ by simply extending a horizontal line from $P=b(\cos t,\sin t)$ and marking its intersection with the ellipse - this yields the point $L$ shown above.
Now we draw a tangent through $L$ and find that its slope is, as expected, $-1$. The "translation" of this line will be its y-intercept, marked as point $M$ in the diagram below. To verify that the translation is equal to $\sqrt{a^2+b^2}$ we can draw a circle through the origin at radius $M$, yielding this:

The circle, as you can see, passes exactly through the point $G$! The radius of this last circle must then be the length of the hypotenuse on our triangle $\Delta HFG$. $$\text{y-intercept}=r=\sqrt{a^2+b^2}$$
Not a proof, but this should give you a more geometric sense of what's going on.
