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  $\frac{X - x}{f_x} = \frac{Y - y }{f_y}$  

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

    Question 1: Find the equation of the tangent at any point of the parabola whose equation is  $y^2 = 4 ax$  
Solution: From the equation of the curve we have,

  $$\begin{align*} y^2 & = 4ax \\ y = & = 2 \sqrt{ax} \\ \therefore \frac{dy}{dx} & = 2 \sqrt{a} \frac{1}{2\sqrt{x}} \\ \frac{dy}{dx} & = \frac{\sqrt{a}}{\sqrt{x}} \\ & = \sqrt{ \frac{a}{\frac{y^2}{4a}}} ~~~~~~~~\bigg| [y^2 = 4ax] \\ & = \sqrt{ \frac{4 a^2}{y^2}} \\ \frac{dy}{dx} & = \frac{2a}{y} \end{align*}$$  

Hence the equation of the tangent is, 
  $$\begin{align*} & Y - y = \frac{2a}{y} (X - x) \\ or, ~ & Yy - y^2 = 2aX - 2ax \\ or, ~ & Yy - 4ax = 2aX - 2ax \\ or, ~ & Yy = 2aX + 2ax \\ or, ~ & Yy = 2a(X + x) \end{align*}$$

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    Question 2: Find the equation of the tangent at any point $(x,y)$ of the ellipse  $\frac{x^2}{a^2} + \frac{ y^2}{b^2} = 1$                
Solution: The equation of the given curve is  $\frac{x^2}{a^2} + \frac{ y^2}{b^2} - 1 = 0$                            

Thus in this case  $f(x,y) = \frac{x^2}{a^2} + \frac{ y^2}{b^2} - 1$  

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
  $$\begin{align*} f(x,y,z) = & \frac{x^2}{a^2} + \frac{ y^2}{b^2} - z^2\\ \textrm{so that,}~~ f_x = & \frac{2x}{a^2}\\ f_y = & \frac{2y}{b^2} \\ f_z = & - 2z \end{align*}$$  
Substituting the values of $f_x, f_y ~\textrm{and}~f_z$ in the symmetrical form equation of the tangent, we have
  $$\begin{align*} & X.\frac{2x}{a^2 } + Y.\frac{2y}{b^2} + Z(- 2z) = 0 \\ or, ~ & \frac{Xx}{a^2} + \frac{Yy}{b^2} = 1 \end{align*}$$ Which is the equation of the tangent ( on replacing Z and z by unity)

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    Question 3: Find the equation of the tangent at any point $(x,y)$ of the curve  $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ .
Solution: The given equation of the curve can be written as  $x^{\frac{2}{3}} + y^{\frac{2}{3}} - a^{\frac{2}{3}} = 0$ which when made homogeneous $x,~y, ~\textrm{and}~z$ becomes  $x^{\frac{2}{3}} + y^{\frac{2}{3}} - a^{\frac{2}{3}} z^{\frac{2}{3}} = 0$  
Thus  $$\begin{align*} f(x,y,z) = & x^{\frac{2}{3}} + y^{\frac{2}{3}} - a^{\frac{2}{3}}~z^{\frac{2}{3}} \\ \therefore f_x = & \frac{2}{3}x^{- \frac{1}{3}} \\ f_y = & \frac{2}{3}y^{- \frac{1}{3}} \\ f_z = & - \frac{2}{3} = a^{\frac{2}{3} z^{- \frac{1}{3}}} \end{align*}$$  
Hence the required equation of the tangent is 
  $$\begin{align*} & X. \frac{2}{3}x^{- \frac{1}{3}} + Y.\frac{2}{3} y^{- \frac{1}{3}} - Z\left( - \frac{2}{3} a^{\frac{2}{3}} z^{- \frac{1}{3}} = 0 \right) \\ or, ~ & \frac{X}{x^{\frac{1}{3}}} + \frac{Y}{y^{\frac{1}{3}}} = a^\frac{2}{3} ~~~\textrm{(replacing Z & z by unity and transposing)} \end{align*}$$

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    Question 4: Find the equation of the tangent at the point 't' on a curve given by  $x = a\cos^3t,~~ y = a\sin^3 t$            

Solution: Here  $$\begin{align*} x = & a \cos^2t \\ \frac{dx}{dt} = & - 3a\cos^2 t \sin t \end{align*}$$  

Also,  $$\begin{align*} y = & a \sin^3t\\ \frac{dy}{dt} = & 3a\sin^2t \cos t \end{align*}$$  

now  $$\begin{align*} \frac{dy}{dx} = & \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ = & - \frac{3a\sin^2 t \cos t}{3a\cos^2t \sin t} \\ = & - \frac{\sin t}{\cos t} \end{align*}$$  

Substituting the values of  $x,~y ~\textrm{and}~ \frac{dy}{dx}$ in the equation,
  $$\begin{align*} Y - y = & \frac{dy}{dx} (X - x) ~\textrm{we get} \\ Y - a \sin^3 t = & - \frac{\sin t }{\cos t} (X - a \cos^3 t) \\ Y \cos t - a \sin^3 t \cos t = & - X \sin t + a \sin t \cos^3 t \\ X\sin t + Y\cos t = & a \sin t \cos^t + a \sin^3 t \cos t \\ = & a \sin t \cos t (\sin^2t + \cos^2t) \\ = & a \sin t\cos t \\ = & a \sin t \cos t\\ \therefore & X \sin t + Y \cos t = a \sin t \cos t \end{align*}$$  

which is the equation of the tangent at the point.