Show that the locus of the foot of the perpendicular on a varying tangent to the ellipse(standard ellipse) from either of its foci is a concentric circle.

A tgt of slope m to to the ellipse (x²/a²) + (y²/b²) = 1, a > b,
centred at the origin O, is : y = mx - v(a²m²+b²), i.e.,
mx - 1y = s, where s = v(a²m²+b²) .................................. (1)
A perpendicular to this tgt, drawn from the focus S(ae,0)
has equation : y = (-1/m)( x - ae ), i.e.,
1x + my = ae ........................................… (2)
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Solving (1) and (2) by Cramer's Rule,
(m²+1)x = (ms+ae) ... and ... (m²+1)y = (aem-s).
Squaring and adding,
(m²+1)² (x²+y²) = [ m²s² + (a²e²) + 2aems ] + [ a²e²m² + s² - 2aems ]
. . . . . . . . . . . . = (m²+1)s² + a²e²(1+m²)
. . . . . . . . . . . . = (m²+1)( s² + a²e² )
. . . . . . . . . . . . = (m²+1)[ (a²m²+b²) + (a²-b²) ], ... where a > b
. . . . . . . . . . . . = (m²+1)² a²
Hence, cancelling (m²+1)², the eq of locus is : x² + y² = a².
This is a Circle with centre O, i.e., concentric with the ellipse. ...... Q.E.D.
In fact, this is the Auxillary Circle of the given ellipse.
Thanks & Regards
Mukesh Sharma
AskIITians Faculty

Approved