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=at2,y=2atx=at^2,y=2at


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

X(t)=x(t)y(t)(x(t)2+y(t)2)x(t)y(t)x(t)y(t)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)+x(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=2ax'=2at,x''=2a

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


X(t)=at22a(4a2t2+4a2)4a2=3at2+2aX(t)=at^2-\frac{2a(4a^2t^2+4a^2)}{-4a^2}=3at^2+2a


Y(t)=2at+2at(4a2t2+4a2)4a2=2at3Y(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±a2m2by = mx±\sqrt{a^2m^2 − b}


f(x,y,m)=ymx±a2m2b=0f(x,y,m)= y - mx±\sqrt{a^2m^2 − b}=0

fm(x,y,m)=x±m/a2m2b=0f'_m(x,y,m)=-x±m/\sqrt{a^2m^2 − b}=0


±a2m2b=mxy±\sqrt{a^2m^2 − b}=mx-y

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

xymx2+m=0xy-mx^2+m=0

m=xyx21m=\frac{xy}{x^2-1}


(yx2yx21)2=a2x2y2(x21)2b(y-\frac{x^2y}{x^2-1})^2=\frac{a^2x^2y^2}{(x^2-1)^2}-b


6.

x/a+y/b=1x/ a + y/ b = 1

f(x,y,a)=x/a+y/b1=0f(x,y,a)=x/a+y/b-1=0

b=1ab=1-a

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

fa(x,y,a)=x/a2+y/(1a)2=0f'_a(x,y,a)=-x/a^2+y/(1-a)^2=0

x(1a)2=ya2x(1-a)^2=ya^2


x/a+x(1+a)/a21=0x/a+x(1+a)/a^2-1=0

x/(1b)+x(2b)/(b1)21=0x/(1-b)+x(2-b)/(b-1)^2-1=0

x(b1)+x(2b)(b1)2=0-x(b-1)+x(2-b)-(b-1)^2=0

x(32b)=(b1)2x(3-2b)=(b-1)^2


7.

x2/a2+y2/b2=1x^ 2/ a^2 + y^ 2/ b^ 2 = 1

x=acost,y=bsintx=acost,y=bsint


evolute of a curve:

X(t)=x(t)y(t)(x(t)2+y(t)2)x(t)y(t)x(t)y(t)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)+x(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)=acostbcost(b2cos2t+a2sin2t)absin2t+abcos2t=acostcost(b2cos2t+a2sin2t)aX(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)=bsintasint(b2cos2t+a2sin2t)absin2t+abcos2t=bsintsint(b2cos2t+a2sin2t)bY(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):

(yy1)=y(x1)(xx1)(y-y_1)=-y'(x_1)(x-x_1)

a2x/x1b2y/y1=a2b2a^2x/x_1 - b^2y/y_1 = a^2 - b^2



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