Question

find envelope of the family of straight line xcosalpha+ysinalpha = l sinalphacosalpha where the parameter is alpha​

Answer 1

Answer:

straight line of sinalphacosalpha

Answer 2

Answer:

√x - √y + √y = 0

CONCEPT- the straight line of cos alpha and sin alpha

FIND - envelope of the family

Step-by-step explanation:

y sin aplha - x sin alpha= a- a sin alpha log tan

ycos alpha/a - xsin alpha/ b = 1

bx+ ay =ab (c-x)xb

ay= a(c-a)

a^2 + a(y -x -c) +cx= 0 (a+b+ c) ....(i)

from using equation first ii we have

2a+y-x-c =0 a= x-y+c /2 .......(ii)

(x+y+z)^2/2 + (x-y+c)/2 (y-x-c) + cx =0

( x-y+c)^2 -4cx= 0

x-y+c -y =0

(√x not equal to √c)^2 -y =0

(√x +√y - √c ) =0 or

√x - √y + √c = 0 is the required equation of the envelope

FINAL ANSWER- √x -√ y + √c = O