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