# Dip of the horizon line onto ellipse behind the tangency point

I have an ellipse of semi-axis major a along x-axis and semi-axis minor b along y-axis.

Having a point C defined with its geodetic (not geocentric) latitude $$\varphi_1$$ on the ellipse surface then drawing a point D along the normal line by height h, so segment CD = h.

Then I draw the tangent lines from D to the ellipse, they touch the ellipse at tangent points F and E respectively.

Another point G, also defined by its geodetic latitude $$\varphi_2$$. The normal line at point G intersects the tangent line (DE) at point H.

My question is, how to find the height m of the segment [GH]? For a circle I would know how to calculate it but for an ellipse, it's complicated :(

Illustration:

• This has nothing whatsoever to do with elliptic-curves. Basically because elliptic curves have relatively little to do with basic properties of conic sections called ellipses. Read the tag descriptions before using them. If that description is all Greek to you, it is a high probability that the tag is inappropriate for your question. – Jyrki Lahtonen Nov 16 '19 at 11:02
• You should explain what you call geodetic altitude : is it the parameter $\alpha$ such that $C=(x,y)=(a \cos \alpha,b \sin \alpha)$ ? – Jean Marie Nov 17 '19 at 1:26
• it is the angle between the normal line of a point on ellipse and the x-axis. what you wrote, $alpha$, I would call it the geocentric latitude because the line containing $O$ point of origin and your point $C$ is always passing by the center of the ellipse. – Khaled Nov 17 '19 at 6:53
• Problem statement is: Given $( \varphi_1, \varphi_2, a,b, h)$ to find $GH=m$ subject to $DH$ being tangential to ellipse. Right? – Narasimham Nov 17 '19 at 21:12
• Yes, exactly! with what you described – Khaled Nov 17 '19 at 21:18

It is cumbersome algebraic work. But I shall indicate the method for you to complete.

At first solve together standard form and its derivative for coords $${C ,G. }$$

Let $$t1= \tan \varphi_1 \,; t2= \tan \varphi_2\,\tag1$$ $$\frac{x_1^2}{a^2} +\frac{y_1^2}{b^2}=1;\, \frac{x_1}{a^2} +\frac{y_1 t2}{b^2}=0;\,\frac{x_2^2}{a^2} +\frac{y_2^2}{b^2}=t2 ;\, \frac{x_2}{a^2} +\frac{y_2 t2}{b^2}=0; \tag2$$

to get points on ellipse $$(C,G)$$

$$x_1= \frac{a^2\, t1}{\sqrt{b^2+a^2t1}}, \,y_1= \frac{b^2}{\sqrt{b^2+a^2t1}}\,;\tag3$$

$$x_2= \frac{a^2\, t2}{\sqrt{b^2+a^2t2}}, \,y_2= \frac{b^2}{\sqrt{b^2+a^2t2}}\,;\tag4$$

Next coordinatess of $$D,H$$ as outward extensions along each normal

$$X_1=x_1 + h \cos \varphi_1; Y_1=y_1 + h \sin \varphi_1; \tag5$$

$$X_2=x_2 +m \cos \varphi_2; Y_2=y_2 +m \sin \varphi_2; \tag6$$

Equation of line joining $$DH$$

$$\dfrac{y-Y_1}{x-X_1}=\dfrac{Y_2-Y_1}{X_2-X_1} \tag7$$

Put the above into $$Ax+By+C=0$$ form.

First find condition that $$Ax+By+ C=0$$ has tangency w.r.t. the standard ellipse.

The discriminant $$\Delta$$ of

$$\dfrac{x^2}{a^2} +\dfrac{1}{b^2}[(C+Ax)/B]^2 -1=0 \tag8$$

Or, ( check this up!)

$$(CA/b^2B^2)^2 = (1/a^2 +A^2/b^2B^2) *[( C^2-b^2B^2)/b^2 B^2] \tag9$$

After checking the above take it forward by simplifying it to find the required formula/condition. It would be implicit in $$m$$ ..

Verify $$m$$ value found out on your Geogebra? graph with numerical values you gave in to make it.

• thank you, I solved that with other method, maybe longer but simpler for me. anyway I didnt test your formulae, I will do it soon, thank you – Khaled Nov 21 '19 at 21:27
• Please show other simpler method too. – Narasimham Nov 22 '19 at 3:33