Answer to Question #281787 in Differential Geometry | Topology for chaitu

Question

4. Find the evolute of the rectangular hyperbola y

2 = 4ax. Ans: 27ay2 = 4(x −

2a)

3

Hint: Take P (at2

, 2at) be any point on the parabola.

5. Find the envelope of the family of lines of the form y = mx±

√︀

a2m2 − b

2. Ans:

x

2

a2

−

y

2

b

2

= 1

6. Find the envelope of the family of lines of the form

x

a

+

y

b

= 1 subject to the

condition a + b = 1. Ans: √

x +

√

b = 1

7. Find the evolute of

x

2

a2

+

y

2

b

2

= 1 as the envelope of the normals. Ans: (ax)

2/3 +

(by)

2/3 = (a

2 − b

2

)

2/3

Answer 1

4.

 $x=at^2,y=2at$

evolute of a curve is the locus of all its centers of curvature:

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

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

$x'=2at,x''=2a$

$y'=2a,y''=0$

$X(t)=at^2-\frac{2a(4a^2t^2+4a^2)}{-4a^2}=3at^2+2a$

$Y(t)=2at+\frac{2at(4a^2t^2+4a^2)}{-4a^2}=-2at^3$

5.

The envelope of the family of curves is a curve such that at each point it touches tangentially one of the curves of the family.

$y = mx±\sqrt{a^2m^2 − b}$

$f(x,y,m)= y - mx±\sqrt{a^2m^2 − b}=0$

$f'_m(x,y,m)=-x±m/\sqrt{a^2m^2 − b}=0$

$±\sqrt{a^2m^2 − b}=mx-y$

$-x+\frac{m}{mx-y}=0$

$xy-mx^2+m=0$

$m=\frac{xy}{x^2-1}$

$(y-\frac{x^2y}{x^2-1})^2=\frac{a^2x^2y^2}{(x^2-1)^2}-b$

6.

$x/ a + y/ b = 1$

$f(x,y,a)=x/a+y/b-1=0$

$b=1-a$

$f(x,y,a)=x/a+y/(1-a)-1=0$

$f'_a(x,y,a)=-x/a^2+y/(1-a)^2=0$

$x(1-a)^2=ya^2$

$x/a+x(1+a)/a^2-1=0$

$x/(1-b)+x(2-b)/(b-1)^2-1=0$

$-x(b-1)+x(2-b)-(b-1)^2=0$

$x(3-2b)=(b-1)^2$

7.

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

$x=acost,y=bsint$

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):

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

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