How do I find the acute angle between two tangents which can be drawn from the point $(1,1)$ to the ellipse $4x^2+9y^2=1$ ? (example)

or simply say, how do I find the angle between two tangents which can be drawn from the point $(x,y)$ to the ellipse $\frac{x^2}{a} + \frac{y^2}{b} = 1$ ?

A graphical representation would be much appreciated, thanks !

  • $\begingroup$ have you compueted the equations of the tangent lines? $\endgroup$ – Dr. Sonnhard Graubner Sep 19 '16 at 15:19
  • $\begingroup$ Its easy ... Just write the equation of a pair of tangents ie ,$$ SS' = T^2 $$ from there you would get an equation for the pair of tangent and from there onward you can find the angle between the tangents very easily . $\endgroup$ – Tejus Sep 19 '16 at 15:23
  • $\begingroup$ How can I do it, without using the formula ? $\endgroup$ – warman Sep 19 '16 at 15:25
  • $\begingroup$ @Dr. Sonnhard Graubner the tangent lines are y=2.1x-1.1 and y=0.56x+0.44 respectively, I guess. What should I do next, sir ? $\endgroup$ – warman Sep 19 '16 at 15:35

We are going to solve this problem in the general case by considering the ellipse as a parametric curve.

The ellipse defined by the equation $$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1$$ is represented parametrically as $$z:(a\cos t,b\sin t)$$ and the derivative vector of the ellipse at some parameter $t$ is $$z':(-a\sin t,b\cos t)$$ Now given a point $p=(p_x,p_y)$ and some point on the ellipse $z(t)$, the line between $p$ and $z(t)$ is a tangent to the ellipse if it is parallel to the derivative vector. In other words, the "planar cross product" $(a,b)\times(c,d)=ad-bc$ of $z(t)-p$ and $z'(t)$ has to be zero: $$(a\cos t-p_x)(b\cos t)-(b\sin t-p_y)(-a\sin t)=0$$ $$ab\cos^2t-p_xb\cos t+ab\sin^2t-p_ya\sin t=0$$ $$ab=p_xb\cos t+p_ya\sin t$$ $$\frac{p_x}a\cos t+\frac{p_y}b\sin t=1$$ This last equation has two solutions for $t$ if $p$ is outside (not on) the ellipse; call these solutions $t_1$ and $t_2$. These correspond to two points on the ellipse $z_1$ and $z_2$, and $pz_1$ and $pz_2$ are the two tangents to the ellipse from $p$. Then the angle between the tangents can be computed from their dot product: $$\theta=\cos^{-1}\frac{(z_1-p)\cdot(z_2-p)}{|z_1-p||z_2-p|}$$

For the given ellipse of $4x^2+9y^2=1$ ($a=\frac12,b=\frac13$) and $p=(1,1)$, $t=2\tan^{-1}(1\pm\frac2{\sqrt3})$. The slopes of the two tangents are $\frac49(3\pm\sqrt3)$, and the angle between them works out to be $$\cos^{-1}\frac{59}{\sqrt{5209}}$$ or approximately 35.2 degrees. Below is a picture of this ellipse and the tangents to it.

Desmos plot


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