Math, asked by sukhmeet319, 7 hours ago

Find the envelope of the family of straight lines y+tx=2at +at^2 , the parameter being t.

Answers

Answered by ktdilaxi1998gmailcom
0

Answer:

t = ( (x-2a ) + √(2a-x )²-4ay) / 2a

or

t = ( (x-2a ) - √(2a-x )²-4ay) / 2a

Step-by-step explanation:

at^2 + 2at- tx -y =0

at^2 +(2a-x) -y = 0

base on Quadratic formula , (x = −b ± √b²-4ac/2a)

so , t = ( - ( 2a-x ) ± √(2a-x )²-4ay) / 2a

t = ( (x-2a ) + √(2a-x )²-4ay) / 2a

or

t = ( (x-2a ) - √(2a-x )²-4ay) / 2a

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