Answer to Question #281790 in Differential Geometry | Topology for chaitu

Question

Find the evolute of

x

2

a2

−

y

2

b

2

= 1 as the envelope of the normals.

Answer 1

$x^ 2/ a^2 + y^ 2/ b^ 2 = 1$

$x=acost,y=bsint$

evolute of a curve:

$X(t)=x(t)-\frac{y'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)}$

$Y(t)=y(t)+\frac{x'(t)(x'(t)^2+y'(t)^2)}{x'(t)y''(t)-x''(t)y'(t)}$

$X(t)=acost-\frac{bcost(b^2cos^2t+a^2sin^2t)}{absin^2t+abcos^2t}=acost-\frac{cost(b^2cos^2t+a^2sin^2t)}{a}$

$Y(t)=bsint-\frac{asint(b^2cos^2t+a^2sin^2t)}{absin^2t+abcos^2t}=bsint-\frac{sint(b^2cos^2t+a^2sin^2t)}{b}$

equation of normals ar point (x₁,y₁):

$(y-y_1)=-y'(x_1)(x-x_1)$

$a^2x/x_1 - b^2y/y_1 = a^2 - b^2$

then:

$a^2x/(acost) - b^2y/(bsint) = a^2 - b^2$

differentiate respect to t:

$xcost+ysint=(a^2-b^2)cos2t$