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Grade: 12th pass
        
Please tell me Q.No.1 & 2  Solution in full detail.
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6 months ago

Answers : (2)

Arun
23742 Points
							
Dear student
 
1.)
 
6 y^2 y’ = 2ax + 3 x^2
 
hence at (a,a)
y’ = 5/6
 
Now tangent
(y – a) = 5/6 (x – a)
 
now put x = 0 and y = 0 sperately
 
then you will get answer as 30 and – 30
6 months ago
Vikas TU
10337 Points
							
Dear student 
Q2. Slope of tangent = dy/dx 
dy/dx = dy/dt / dx/dt 
dx/dt , 
x = t^2+3t-8
dx/dt = 2t +3 
y = 2t^2 -2t-5 
dy/dt = 4t-2 
dy/dx = d/dt /x/dt 
dy/dx = 4t-2/2t+3 
Put the point (2,-1)
 x = t^2 +3t-8 
2 = t^2 +3t-8 
t =2 and t= -5 
y = 2t^2-2t -5 
-1 = 2t^2 -2t-5 
t = -1 and 2 
t=2 is common in both part 
calculate dy/dx at t = 2 
dy/dx = 6/7 
6 months ago
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