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Given the angle between the tangent and the line that connects the point of tangency to each foci, and you are given the distance from one of the foci to the point of tangency. You are given two angles and two distances can you calculate the length of the major axis?

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  • $\begingroup$ If you dont understand the question just say so and Ill try to help clarify $\endgroup$ – user5200731 Dec 2 '15 at 0:10
  • $\begingroup$ what is the second distance ? $\endgroup$ – WW1 Dec 2 '15 at 0:18
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I have a feeling that the problem has not been stated completely. However, it is interesting that a solution still exists.

Without loss of generality assume $a$ is the major axis aligned with the $x$ axis and the center is at the origin.

Let $l_1$, $l_2$ be the lengths of the lines from the foci to the point $(x,y)$ at which the tangent touches the ellipse. We know that $l_1 + l_2 = 2a$. So if $l_1$ is known we just need to find $l_2$.

Let $\theta_1, \theta_2$ be the angles made by lines from the foci to the point $(x,y)$ (lengths $l_1$, $l_2$) to the $x$-axis: $$ \sin\theta_1 = \frac{y}{l_1} ~,~~ \sin\theta_2 = \frac{y}{l_2} \,. $$

Let $\alpha_1, \alpha_2$ be the angles made by the tangent to the lines of length $l_1, l_2$. Then, $$ \theta_1 + \theta_2 = \alpha_1 + \alpha_2 \,. $$ Therefore, $$ \begin{aligned} l_2 &= \frac{y}{\sin\theta_2} \\ & = \frac{y}{\sin(\alpha_1+\alpha_2-\theta_1)} \\ & = \frac{l_1\sin\theta_1}{\sin(\alpha_1+\alpha_2-\theta_1)} \end{aligned} $$ We still don't know $\theta_1$. Consider the case where $$ \theta_1 = \alpha_1 + \theta $$ where $\theta$ is the angle made by the tangent to the $x$-axis.

The equation of the ellipse is $$ y^2 = b^2\left(1 - \frac{x^2}{a^2}\right) $$ The slope of the tangent is (from the $x$-axis) $$ \tan\theta = \frac{dy}{dx} = -\frac{b^2\,x}{a^2\,y} $$ Therefore $\theta$ is a function of $y$, $\theta = f(y)$. We already know that $\theta_1$ is a function of $y$, $\theta_1 = g(y)$. We can then write an equation for $y$, $$ g(y) = \alpha_1 + f(y) $$ and solve for $y$ and from the solution compute $\theta_1$ followed by $l_2$ and finally $a$. I can't find a closed form solution, but one can solve for $y$ numerically.

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Use the Cosine rule to work out the third length (from the second foci to the point of tangency) then the semi-major radius is just half of the sum of the two lengths to the point of tangency.

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