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A person P moves such that projection of his distance $ D$ to a fixed point C onto a fixed line L through C is proportional to distance left after removing constant length $L$ from C. Find his path.

EDIT1

It may be of interest to note that 1/R is eccentricity and L is latus rectum for all conics in polar form. And the question presents another new definition of conics without directrix.

EDIT2:

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A quick sketch on Geogebra verifies that ratio of (projected segment length on x-axis) to (focal ray length minus latus rectum segment length) equals to chosen eccentricity $\epsilon$ of three constructed conics.

$$ \dfrac{r-p}{x}= \epsilon \quad ,\frac{EE_d}{Oe} = \epsilon_{ell},\, \frac{PP_d}{Op} = \epsilon_{par},\, \frac{HH_d}{Oh} = \epsilon_{hyp}.\, $$

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    I think is a hyperbola. I write now my reasoning. – 2017-02-12

1 Answers 1

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Without loss of generality, let be L the polar axis, the fixed point C the origin and the coordinates of that person $P=(r,\theta)$. So, $D=r$. The projection onto L is $D\cos\theta$. $D-L$ is the diminished distance. So.

$$\frac{D\cos\theta}{D-L}=R$$

With $R$ the constant of proportionality.

$$D=\frac{RL}{R-\cos\theta}=\frac{L}{1-(1/R)\cos\theta}$$

It's an ellipse or a hyperbola depending on the value of the parameters.

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    How to subtract polar axis? – 2017-02-12
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    I understand that the constant $L$ is substracted to the distance $D$. Is it correct? – 2017-02-12
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    Yes But in the first line it is written it as an axis. Grammar? – 2017-02-12
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    Not sure about what you mean. L is the polar axis $L$ is the constant substracted. – 2017-02-12
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    OK, now got it, L is axis and $L$ is the line segment.Fine.. – 2017-02-12