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See illustration I know point $P_1 = (x_1,y_1)$ which sits on a line $L_1$. $P_1$ will lie on the semi-minor axis. $L_1$ is parallel to the semi-major axis. Have another line $L_2$ which is tangent to the ellipse elsewhere in the first or fourth quadrants. I know a point $P_2 = (x_2,y_2)$ on this line which should also lie on the ellipse.Ellipse Problem

$P_1$, $P_2$, $L_1$, $L_2$ are all known. My problem is how do I find $a$ and $b$, the length of the semi-major and semi-minor axes? My intuition tells me my problem is constrained enough to be solved to a single ellipse.

I started with the Cartestian equation for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, did some implicit differentiation to get $\frac{dy}{dx}=\frac{-xb^2}{ya^2}$. I know the slope of the tangent line $L_2$, so thought this would be the best line of attack but have been staring at this ages now and don't know how to proceed. What I'm looking to get is two linear equations with $a$ and $b$ in them, right?

I went down the path of rearranging the derivative equation. I've substituted $m_2$ for $\frac{dy}{dx}$; the slope of $L_2$:

$$a = \sqrt{\frac{-x_2b^2}{y_2m_2}}$$

but don't see how this helps me because it's not a linear equation. How can I solve for my ellipse?

EDIT: I used the equation of an ellipse centred at the origin above, but this was a mistake. I'm happy to rotate/translate lines to make the calculations simpler and then transform back e.g. so that $x_1 = 0$.

PeteUK
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  • If you can upload an illustration it would be helpful! Thanks! – jonsno Jun 19 '17 at 08:56
  • @samjoe I was too eager with the "Post" button. Illustration now uploaded. – PeteUK Jun 19 '17 at 08:57
  • Do you know all $P_1, P_2, L_1, L_2$ ? – jonsno Jun 19 '17 at 09:15
  • @samjoe yes, they're known. – PeteUK Jun 19 '17 at 09:24
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    The form of your ellipse's equation implies that the center is at $(0,0)$ and $|y_1|=b$, $x_1=0$. If that is not given, you have to modify the ellipse's equation. – ccorn Jun 19 '17 at 09:28
  • @ccorn That's a good point. The ellipse isn't centred at the origin. – PeteUK Jun 19 '17 at 09:41
  • Also $m_1=0$ in that setting. If that is not the case, you'll have to rotate the ellipse as well, or use the general algebraic form of the conic section equation. – ccorn Jun 19 '17 at 09:43
  • @ccorn To simplify things I was going to rotate the lines to make $L_1$ parallel to the x-axis, solve, then rotate back. – PeteUK Jun 19 '17 at 09:47
  • OK, then $m_1=0$. Have you also translated such that $x_1=0$? Then you only need a $y_0$ coordinate for the center. – ccorn Jun 19 '17 at 09:49
  • @ccorn I hadn't thought about that, but could do the translation so $x_1 = 0$ – PeteUK Jun 19 '17 at 09:52

2 Answers2

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Since you do not actually have the ellipse's center given, you need a more general form for the ellipse's equation.

Here is a geometric hint; I leave it to you to translate this to algebra.

You know the minor symmetry axis of the ellipse: It is the line orthogonal to $L_1$ passing through $P_1$. Now all you need to determine the center $O$ is another non-parallel line known to pass through $O$. Let $M$ be the midpoint between $P_1$ and $P_2$. Let $Q$ be the intersection point of $L_1$ and $L_2$. Then the straight line $MQ$ passes through $O$.

Ellipse with tangents

Why? Because that would be the case if the ellipse were a circle. Deforming a circle into an arbitrary ellipse can be done with an invertible affine transformation. Such transformations preserve line straightness, tangency and midpoint relationships. Therefore $MQ$ still passes through $O$ after the transformation.

ccorn
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  • Plugged this in and works a treat :) I like your reasoning about the transformation preserving midpoint relationships, etc. BTW, did you use a particular program to generate your drawing? I would like an alternative to uploading images. – PeteUK Jun 19 '17 at 21:01
  • @PeteUK: The figure has been generated with an older, Java-based, desktop version of Geogebra. – ccorn Jun 19 '17 at 23:39
  • Nice answer (+1). – Hypergeometricx Jun 20 '17 at 09:14
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Assume that $P_1=(0,b)$ and $P_2=(a\cos\theta, b\sin\theta)=(h,k)$, both lying on the ellipse $\frac {x^2}{a^2}+\frac{y^2}{b^2}=1$, and that $L_1: y=b$, and $L_2: x+By+C=0\tag{1}$.

Quantities $b,h,k,B,C$ are known. We only need to find the value of $a$ to determine the ellipse.

First we can determine $\theta$ as equating $k=b\sin\theta$, and consequently determine a using $h=a\cos\theta$.

By differentiating and using standard formulas, we find that the equation of the tangent at $P$ is

$$\frac x{a\sin\theta}+\frac y{b\cos\theta}-(\tan\theta+\cot\theta)=0$$ which can be written as

$$x+\left(\frac ab\tan\theta\right)y-a\sin\theta(\tan\theta+\cot\theta)=0\tag{2} $$

Using values of $a,\theta$ as computed, coefficients in equations $(1),(2)$ should turn out to be equal.

Hypergeometricx
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  • Does this answer assume the ellipse is centred at the origin? It isn't. I've updated my answer to show that. Doesn't that now mean that $h$ and $k$ are unknown? Is $C$ also unknown in equation of $L_2$? Is equation of $L_2$ missing the coefficient on $x$? How come $P_1 = (0,-b)$? Shouldn't it's y-value by positive if it's above the ellipse's centre? – PeteUK Jun 19 '17 at 11:43
  • It assumes that the method applies after you have done the translation such that $x_1=0$. $h,k$ are known as question specifies both points are known. Both $B, C$ are known as question specifies both lines are known. No coeff for $x$, as other coeffs assumed to be normalised. Works for both $P_1=(0,-b), L_1: y=-b$ as well as $P_1=(0,b), L_1: y=b$. Now changed to the latter for greater clarity. – Hypergeometricx Jun 19 '17 at 14:20