MATH 121 (Wg Cdr Monir)
Tangent and Normal
(Related Problems)
Some points to remember:
• If the equation of the curve is f(x,y)=0f(x,y)=0 the equation of the normal at (x,y)(x,y) is X−xfx=Y−yfyX−xfx=Y−yfy
• If the equation of the curve is x=ϕ(t), y=ϕ(t)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==2√ax∴dydx=2√a12√xdydx=√a√x=√ay24a |[y2=4ax]=√4a2y2dydx=2ay
Hence the equation of the tangent is,
Y−y=2ay(X−x)or, Yy−y2=2aX−2axor, Yy−4ax=2aX−2axor, 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+y2b2−1=0
Thus in this case f(x,y)=x2a2+y2b2−1
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+y2b2−z2so 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+y23−a23=0
which when made homogeneous x, y, and z becomes
x23+y23−a23z23=0
Thus
f(x,y,z)=x23+y23−a23 z23∴fx=23x−13fy=23y−13fz=−23=a23z−13
Hence the required equation of the tangent is
X.23x−13+Y.23y−13−Z(−23a23z−13=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,
Y−y=dydx(X−x) we getY−asin3t=−sintcost(X−acos3t)Ycost−asin3tcost=−Xsint+asintcos3tXsint+Ycost=asintcost+asin3tcost=asintcost(sin2t+cos2t)=asintcost=asintcost∴Xsint+Ycost=asintcost
which is the equation of the tangent at the point.
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