[Math] Calculate tangent point on ellipse

conic sectionsgeometry

I'm trying to find a tangent point on an ellipse. Trying a lot, using answers found a.o. on this site, but obviously doing something wrong as I'm not getting any good results.

I've added a sketch, to help make it more clear.

The problem at hand:

I have an ellipse, and am basically looking for the $45^\circ$ tangent point, this is point P in the sketch.

The center of the ellipse is $(0,0)$.

$P$ is $200$ to the right from the origin, and $(b-22)$ up from the origin, so $P$ is at $(200, b-22)$.

For a given major radius $a$, I try to calculate the minor radius $b$.

The answer I need is in the form of $a = f(b)$.

How to do this?

(addition)
The formula I managed to derive in different ways before asking here (and which came back in the answers) is $ a = \sqrt{200 \frac{b^2}{b-22}} $. Now when I fill in b=30, I get a=150. That's impossible as in this case point P is not on the ellipse, it's somewhere out of the ellipse.

(solution)
First of all, thanks to all that helped out. In the end I realised I was much closer to the answer than I thought; but needed the final bits to put it all together. I've been drawing some nice ellipses now – those things are remarkably tricky to work with! Furthermore, to allow for the needed flexibility, I've replaced the value 200 for x in the image above, with the variable x.

The correct question is: what are the major and minor radiuses of an ellipse so that it passes through point P = (x,b-22) with slope -1?

Slope -1 at P:
$$a = \sqrt{\frac{xb^2}{b-22}} $$

Point P on the ellipse:
$$\frac{x^2}{a^2}+\frac{(b-22)^2}{b^2} = 1 $$

Substitute the first in the second, and solve:

$$b = \frac{22x-484}{x-44} $$

a in turn is caculated with the first formula.
For more details on how these formulas are derived, see also the replies below.

sketch of the problem at hand

Best Answer

The problem is that the given constraints determine $b$; you are not free to take $b=30$. As others have shown, the constraint on the tangent gives $$ a = b\sqrt{\frac{200}{b-22}} $$ The constraint that $P$ be on the ellipse gives $$ \frac{200^2}{a^2} + \frac{(b-22)^2}{b^2} = 1 $$ Use the first to eliminate $a$ from the second, and you'll get an equation for $b$, which is readily solved to give $$ b = \frac{979}{14} \approx 69.9 $$

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