Find the locus of the feet of the perpendicular drawn upon any tangent to the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ from either focus.

Asked Jan 18 '20 at 12:12

Active Jan 19 '20 at 05:14

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Find the locus of the feet of the perpendicular drawn upon any tangent to the ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ from either focus.

My Attempt: Let the tangent to the ellipse be $y=mx\pm \sqrt {a^2m^2+b^2}$ Here the slope of tangent is $m$ so the slope of the perpendicular drawn from focus to the tangent is $-\dfrac {1}{m}$.

Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ This follows a geometrical approach but I'm looking for an analytical solution.

edited Jan 19 '20 at 05:14

Martin Sleziak

53,687

asked Jan 18 '20 at 12:12

pi-π

7,416

https://math.stackexchange.com/questions/775323/equation-of-the-locus-of-the-foot-of-perpendicular-from-any-focus-upon-any-tange – lab bhattacharjee Jan 18 '20 at 12:32
It's simply axillary circle. – mathsdiscussion.com Jan 18 '20 at 12:36
Does this answer your question? Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ – Andrei Jan 18 '20 at 12:45
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@Andrei The OP explicitly said that question did not help him because it did not give an analytic solution, only a geometric one. – almagest Jan 18 '20 at 13:30
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@almagest Note that the request for analytical approach came after the link to the geometric approach was posted – Andrei Jan 18 '20 at 14:37
Your attempt so far consists of writing down two definitions. That’s a start, but I think you can get farther than that. How about developing the equation of the perpendicular through a focus now? – amd Jan 18 '20 at 19:16
See my answer for the visualization, your required locus is the auxiliary circle (green). – Ng Chung Tak Jan 19 '20 at 08:36

Find the locus of the feet of the perpendicular drawn upon any tangent to the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ from either focus.

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