Latus rectum of conic sections

The following description is from Wikepedia https://en.wikipedia.org/wiki/Parabola#Axis_of_symmetry_parallel_to_the_y_axis (retrieved 10-NOV-2021, 18:14).

The latus rectum is the line drawn through a focus of a conic section parallel to the directrix and terminated both ways by the curve. For any case, the semi-latus rectum is the radius of the osculating circle at the vertex. For a parabola, the semi-latus rectum is the distance of the focus from the directrix.

We shall verify the last statement. Firstly, let us compute the semi-latus for ellipse, hyperbola and parabola. We shall denote the semi-latus (resp. latus) rectum by $p$ (resp. $l$ ).

For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \; (a>b>0)$ , we have $p=b^2/a$ . In the limit case, where $a=b$ and the ellipse is reduced to a circle of radius $a$ , we have $p=a$ .

For a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \; (a, b>0)$ , again we have $p=b^2/a$ .

For a parabola with the origin as its vertice and focal length $f$ , its equation is $x^2=4fy$ (assume that the axis of our parabola is the $y$ -axis). We have $2p=4f$ in this case, that is, $p=2f$ .

Recall that the osculating circle is just the curvature circle, its radius is the curvature radius, which is equal to the reciprocal of the curvature at the point under consideration. Using parametric equation of a regular planar curve $C: x=x(t), y=y(t)$ , the osculating circle radius is $R(t)=\frac{1}{|\kappa(t)|}$ , where the curvature $\kappa(t)$ is given by

$\displaystyle \kappa(t) = \frac{x'(t)y''(t)-x''(t)y'(t)}{\big( x'(t)^2+y'(t)^2\big)^{3/2}}$

So, for the above ellipse, whose parametric equation is $x=a\cos(t), y = b\sin(t)$ , an osculating circle at the right vertex $(a, 0)$ (corresponding to $t=0$ ) has radius

$\displaystyle R(t) = \frac{(a^2\sin^2(t)+b^2\cos^2(t))^{3/2}}{ab}\big\vert_{t=0} = \frac{b^3}{ab}=\frac{b^2}{a}$ , as desired.

Let us check the case of a hyperbola. The parametric equation is $x=a\cosh(t), y=b\sinh(t)$ . An osculating circle at the right vertex $a=0$ (corresponding to $t=0$ ) has radius

$\displaystyle R(t)=\frac{a^2\cosh^2(t)+b^2\sinh^2(t)}{|-ab|}\big\vert_{t=0}=\frac{b^2}{a}$ ,

and hence also correct.

Finally we verify the case of a parabola. The parametric equation is $x=t, y=t^2/2p$ . Basic calculation shows that the osculating circle at the vertex $(0, 0)$ (corresponding to $t=0$ ) has radius $R(t)\vert_{t=0}= p(1+\frac{t^2}{p^2})^{3/2} \big\vert_{t=0} = p$ .

We have thus finished the proof.

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