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Lecture 18 [07 June, 2020 ] MATH 121 (Wg Cdr Monir)

 MATH 121 (Wg Cdr Monir) 

Tangent and Normal

(Related Problems)




Some points to remember:

    • If the equation of the curve is  f(x,y)=0   the equation of the normal at (x,y) is  Xxfx=Yyfy  

    • If the equation of the curve is  x=ϕ(t), y=ϕ(t) , the equation of the normal at the point 't' is  ϕ(t)X+ϕ(t)Y=ϕ(t) ϕ(t)+ϕ(t)+ϕ(t)  


    Question 1: Find the equation of the tangent at any point of the parabola whose equation is  y2=4ax  
Solution: From the equation of the curve we have,

  y2=4axy==2axdydx=2a12xdydx=ax=ay24a        |[y2=4ax]=4a2y2dydx=2ay  

Hence the equation of the tangent is, 
  Yy=2ay(Xx)or, Yyy2=2aX2axor, Yy4ax=2aX2axor, Yy=2aX+2axor, Yy=2a(X+x)


    Question 2: Find the equation of the tangent at any point (x,y) of the ellipse  x2a2+y2b2=1                
Solution: The equation of the given curve is  x2a2+y2b21=0                            

Thus in this case  f(x,y)=x2a2+y2b21  

Which is an algebraic function not homogeneous of the seco& degree in x and y. Making it homogeneous of the second degree by the introduction of the linear unit z, we have
  f(x,y,z)=x2a2+y2b2z2so that,  fx=2xa2fy=2yb2fz=2z  
Substituting the values of fx,fy and fz in the symmetrical form equation of the tangent, we have
 X.2xa2+Y.2yb2+Z(2z)=0or, Xxa2+Yyb2=1 Which is the equation of the tangent ( on replacing Z and z by unity)


    Question 3: Find the equation of the tangent at any point (x,y) of the curve  x23+y23=a23 .
Solution: The given equation of the curve can be written as  x23+y23a23=0 which when made homogeneous x, y, and z becomes  x23+y23a23z23=0  
Thus  f(x,y,z)=x23+y23a23 z23fx=23x13fy=23y13fz=23=a23z13  
Hence the required equation of the tangent is 
  X.23x13+Y.23y13Z(23a23z13=0)or, Xx13+Yy13=a23   (replacing Z & z by unity and transposing)


    Question 4: Find the equation of the tangent at the point 't' on a curve given by  x=acos3t,  y=asin3t            

Solution: Here  x=acos2tdxdt=3acos2tsint  

Also,  y=asin3tdydt=3asin2tcost  

now  dydx=dydtdxdt=3asin2tcost3acos2tsint=sintcost  

Substituting the values of  x, y and dydx in the equation,
  Yy=dydx(Xx) we getYasin3t=sintcost(Xacos3t)Ycostasin3tcost=Xsint+asintcos3tXsint+Ycost=asintcost+asin3tcost=asintcost(sin2t+cos2t)=asintcost=asintcostXsint+Ycost=asintcost  

which is the equation of the tangent at the point.


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