Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + ln(t), y = t2 + 6, (6, 7)

2x-y=5

Step-by-step explanation:

Given

without eliminating parameter

=

=2t

at

6=6+

t=1

Equation of line is given by

2=

2x-12=y-7

2x-y=5

by eliminating parameter

differentiating we get

2x-y=5

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )