Asked by MeGaTrOn
                Find the normal to the curve x=tt+1x=tt+1 and y=tt−1y=tt−1 at the point T where t = 2.
I tried converting this into one equation (y=x(t+1)t−1y=x(t+1)t−1) but don't see how to get the answer which says:
dydx=−(t+1)2(t−1)2dydx=−(t+1)2(t−1)2
and that T=(23,2)T=(23,2), 3x−27y+52=03x−27y+52=0
I'd appreciate any pointers.
            
        I tried converting this into one equation (y=x(t+1)t−1y=x(t+1)t−1) but don't see how to get the answer which says:
dydx=−(t+1)2(t−1)2dydx=−(t+1)2(t−1)2
and that T=(23,2)T=(23,2), 3x−27y+52=03x−27y+52=0
I'd appreciate any pointers.
Answers
                    Answered by
            Steve
            
    all that duplicated text is impossible to parse. Try typing it in, rather than copy/pasting.
Use t^2 for t squared
    
Use t^2 for t squared
                    Answered by
            MeGaTrOn
            
    Find the normal to the curve 
X=t÷[t+1] and Y=t÷[t-1]at point T where t=2.
I tried converting this into one equation Y=x(t+1)÷(t-1) but don't see how to get the answer which says:
dy÷dx=-(t+1)^2÷(t-1)^2
and that T=(2÷3,2), 3x−27y+52=0
Is this ok
Those problems i got are for an pdf i got from my sir
    
X=t÷[t+1] and Y=t÷[t-1]at point T where t=2.
I tried converting this into one equation Y=x(t+1)÷(t-1) but don't see how to get the answer which says:
dy÷dx=-(t+1)^2÷(t-1)^2
and that T=(2÷3,2), 3x−27y+52=0
Is this ok
Those problems i got are for an pdf i got from my sir
                    Answered by
            Steve
            
    All the steps in your work are correct.
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.