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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

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!