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

Question #281787



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


1
Expert's answer
2021-12-22T14:58:10-0500

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\u00b1\\sqrt{a^2m^2 \u2212 b}"


"f(x,y,m)= y - mx\u00b1\\sqrt{a^2m^2 \u2212 b}=0"

"f'_m(x,y,m)=-x\u00b1m\/\\sqrt{a^2m^2 \u2212 b}=0"


"\u00b1\\sqrt{a^2m^2 \u2212 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\/\n\na\n\n+\n\ny\/\n\nb\n\n= 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^\n\n2\/\n\na^2\n\n+\n\ny^\n\n2\/\n\nb^\n\n2\n\n= 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"



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