conics again (1 Viewer)

[sonnylongbottom]

conics is doing my head in.

Show that the locus of the foot of the perpendicular drawn from the origin to the tangent to the curve xy=c^2 at the point P (ct, c/t) is given by (x^2+y^2)^2 = 4c^2xy

 

[Affinity]

method 1:
find the equation of a tangent and the perpendicular, solve, etc. (this is the usual way)

dy/dx = dy/dy / dx/dt = -1/t^2

so the tangent is
(y - c/t) = -(x - ct)/t^2 -------> x + yt^2 =2ct
and the perpendicular has gradient t^2 through the origin, so its formula is
y = t^2 x

solving gives
x = 2ct/(1+t^4)
y = 2ct/(t^2 + 1/t^2)

so

y = 2c[sqrt(y/x)]/[ x/y + y/x]
sqrt(xy) = 2c / [ (x^2 + y^2)/ xy]
x^2 + y^2 = 2c* sqrt(xy)

and squaring gives what you are after.