Eccentric angles of points of contact of two parallel tangents in an ellipse

Question

The following statement is given in my book under the topic Tangents to an Ellipse:

The eccentric angles of the points of contact of two parallel tangents differ by $\pi$

In case of a circle, it is easy for me to visualise that two parallel tangents meet the circle at two points which are apart by $\pi$ radians as they are diametrically opposite. But in case of ellipse, as the eccentric angle is defined with respect to the auxiliary circle and not the ellipse, I am unable to understand why two parallel tangents meet the ellipse at points which differ by $\pi$.

Kindly explain the reason behind this fact.

Answer 1

Kindly explain the reason behind this fact.

The reason is that an ellipse can be obtained by stretching/shrinking a circle. The strech/shrink is a linear map (linear transformation).

Let's consider two tangent lines on the circle $x^2+y^2=a^2$ at $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$. You already know that the two tangent lines are parallel.

Now, let's stretch/shrink the circle and the tangent lines. Stretching/shrinking the circle $x^2+y^2=a^2$ to obtain the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ means that you replace $y$ in $x^2+y^2=a^2$ with $\frac{a}{b}y$ to have $x^2+\left(\frac aby\right)^2=a^2$ which is nothing but $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

By this stretch/shrink, we have the followings :

• The circle $x^2+y^2=a^2$ is transformed to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

• The two parallel lines are transformed to two parallel lines.

• The two lines tangent to the cirlce are transformed to two lines tangent to the ellipse.

• The tangent points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$ on the circle are transformed to two tangent points $(a\cos\theta,b\sin\theta)$,$(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse respectively.

From the above facts, it follows that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$.

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The followings are the proof for the above facts.

Let's consider the circle $x^2+y^2=a^2$ and two points $(a\cos\theta,a\sin\theta)$,$(a\cos(\theta+\pi),a\sin(\theta+\pi))$.

The equation of the tangent line at $(a\cos\theta,a\sin\theta)$ is given by $$a\cos\theta\ x+a\sin\theta\ y=a^2\tag1$$

Similarly, the equation of the tangent line at $(a\cos(\theta+\pi),a\sin(\theta+\pi))$ is given by $$a\cos(\theta+\pi)x+a\sin(\theta+\pi)y=a^2\tag2$$

Now, let's stretch/shrink the circle and the lines $(1)(2)$ by replacing $y$ with $\frac aby$ to have $$(1)\to a\cos\theta\ x+a\sin\theta\cdot\frac aby=a^2\tag3$$ $$(2)\to a\cos(\theta+\pi)x+a\sin(\theta+\pi)\cdot\frac aby=a^2\tag4 $$ Here note that these lines $(3)(4)$ are parallel since the slope of each line is $\frac{-b\cos\theta}{a\sin\theta}$.

Finally, note that $(3)$ can be written as $$\frac{a\cos\theta}{a^2}x+\frac{b\sin\theta}{b^2}y=1\tag5$$ which is nothing but the tangent line at $(a\cos\theta,b\sin\theta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Similarly, $(4)$ can be written as $$\frac{a\cos(\theta+\pi)}{a^2}x+\frac{b\sin(\theta+\pi)}{b^2}y=1\tag6$$ which is nothing but the tangent line at $(a\cos(\theta+\pi),b\sin(\theta+\pi))$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Since $(5)(6)$ are parallel, we see that the eccentric angles of the points of contact of two parallel tangents differ by $\pi$. $\quad\square$

Answer 2

According to the definition of the eccentric angle for the ellipse $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$,

$$ t= \tan^{-1} \frac{ay}{bx}$$

evaluate

$$t_2-t_1= \tan^{-1} \frac{ay_2}{bx_2} - \tan^{-1} \frac{ay_1}{bx_1}=\tan^{-1}\frac { \frac{ay_2}{bx_2} - \frac{ay_1}{bx_1} } {1+ \frac{ay_2}{bx_2} \frac{ay_1}{bx_1} }\tag{1}$$

The tangent of the ellipse is $-\frac{b^2x}{a^2y}$. So, the two parallel tangents satisfy,

$$\frac{x_1}{y_1}=\frac{x_2}{y_2}\tag{2}$$

Plug (2) in to (1),

$$t_2-t_1=\tan^{-1} (0)$$

Thus, the two angles are $\pi$ apart.