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Intersection of line with ellipse - given difference of eccentric angles

Pranav

Well-known member
Nov 4, 2013
428
Problem:

Find the condition so that the line px+qy=r intersects the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in points whose eccentric angles differ by $\frac{\pi}{4}$.

Attempt:

Let the points on ellipse be $(a\cos\theta,b\sin\theta)$ and $(a\cos\left(\frac{\pi}{4}+\theta\right),b \sin\left(\frac{\pi}{4}+\theta\right))$. The slope of line passing through these points is:

$$\frac{b\sin\left( \frac{\pi}{4}+ \theta \right)-b\sin\theta}{a\cos\left(\frac{\pi}{4}+\theta\right)-a\cos\theta}=-\frac{b}{a}\cot\left(\frac{\pi}{8}+\theta\right)$$

The line passing through these two points is:

$$y-b\sin\theta=-\frac{b}{a}\cot\left(\frac{\pi}{8}+\theta\right)(x-a\cos\theta)$$
Next step involves rearranging the equation and comparing the coefficients with that of px+qy=r but trying that leads to some dirty equations and I am inclined to think that there exists a lot better way to solve the given problem. Can anyone give me a few hints about how to proceed?

Any help is appreciated. Thanks!
 

Pranav

Well-known member
Nov 4, 2013
428
Thanks to whoever tried this problem, the problem has been solved. I missed a very obvious way for this problem.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Thanks to whoever tried this problem, the problem has been solved. I missed a very obvious way for this problem.
I wonder what the "very obvious way" is? My method would be to dilate the $y$-axis by a factor $a/b$. Then the ellipse becomes the circle $x^2+y^2 = a^2$ and the line becomes $apx + bqy - ar = 0.$ Angles around the circle are (by definition) the same as eccentric angles for the ellipse. So we want the line to cut the circle at two points whose angles differ by $\pi/4$. The condition for that is that the distance from the line to the origin is $a\cos(\pi/8)$, or $$\frac{|ar|}{\sqrt{a^2p^2 + b^2q^2}} = a\cos(\pi/8).$$ So the condition on $p$, $q$ and $r$ is $|r| = \sqrt{a^2p^2 + b^2q^2}\cos(\pi/8).$
 

Pranav

Well-known member
Nov 4, 2013
428
Hi Opalg! :)

I wonder what the "very obvious way" is? My method would be to dilate the $y$-axis by a factor $a/b$. Then the ellipse becomes the circle $x^2+y^2 = a^2$ and the line becomes $apx + bqy - ar = 0.$ Angles around the circle are (by definition) the same as eccentric angles for the ellipse. So we want the line to cut the circle at two points whose angles differ by $\pi/4$. The condition for that is that the distance from the line to the origin is $a\cos(\pi/8)$, or $$\frac{|ar|}{\sqrt{a^2p^2 + b^2q^2}} = a\cos(\pi/8).$$ So the condition on $p$, $q$ and $r$ is $|r| = \sqrt{a^2p^2 + b^2q^2}\cos(\pi/8).$
Wow, that looks way shorter than my method but I haven't ever seen the dilation of axis, I am interested in learning more about it, can you please share a relevant link? Thanks.

As for my method, I need to use a result that the point of intersection of tangents to ellipse at eccentric angles $\phi$ and $\phi'$ is given by

$$x=a\cfrac{\cos\left(\frac{\phi+\phi'}{2}\right)}{\cos\left(\frac{\phi-\phi'}{2}\right)}, y=b\cfrac{\sin\left(\frac{\phi+\phi'}{2}\right)}{ \cos \left(\frac{\phi-\phi'}{2}\right)}$$

The chord of contact for this intersection point is same as px+qy=r, from here I can compare the coefficients to obtain the answer.

Agreed that the result for intersection is not a nice one but in my course, I am required to memorise it.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I wonder what the "very obvious way" is? My method would be to dilate the $y$-axis by a factor $a/b$. Then the ellipse becomes the circle $x^2+y^2 = a^2$ and the line becomes $apx + bqy - ar = 0.$ Angles around the circle are (by definition) the same as eccentric angles for the ellipse. So we want the line to cut the circle at two points whose angles differ by $\pi/4$. The condition for that is that the distance from the line to the origin is $a\cos(\pi/8)$, or $$\frac{|ar|}{\sqrt{a^2p^2 + b^2q^2}} = a\cos(\pi/8).$$ So the condition on $p$, $q$ and $r$ is $|r| = \sqrt{a^2p^2 + b^2q^2}\cos(\pi/8).$
Additionally, we have the freedom to pick $a^2p^2 + b^2q^2=1$, without losing any lines.
Furthermore, we can leave out the absolute value function.

That is, $(ap, bq)$ is the vector to the (dilated) unit circle that is perpendicular to the line.
And $r$ is the distance of the (dilated) line to the origin.

If we do that, the condition under $p$, $q$, and $r$ becomes:
\begin{cases}
a^2p^2 + b^2q^2=1 \\
r = \cos(\pi / 8)
\end{cases}
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi Pranav,
You asked about dilation along an axis. I've found the following to be very useful.

MHBgeometry6.PNG
 

Pranav

Well-known member
Nov 4, 2013
428
Hi johng! :)

Thank you for taking the time to explain the dilation of axis but I fear I cannot understand that as I am not well versed with the notation you have used. I will see if I can find something on the internet or books. Thank you once again.