Examples on Tangents and Normals Set 3

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Example – 6

Find all the tangents to the curve \(y = \cos \left( {x + y} \right),\,\,\, - 2\pi  \le x \le 2\pi ,\)  that are parallel to the line \(x + 2y = 0\).

Solution: We require the slope of the tangents to be \(\begin{align}\frac{{ - 1}}{2}.\end{align}\)

Differentiating the given equation of the curve, we get:

\[\begin{align}&\frac{{dy}}{{dx}}=  - \sin \left( {x + y} \right)\left\{ {1 + \frac{{dy}}{{dx}}} \right\}\\\\\Rightarrow \qquad \qquad &\frac{{dy}}{{dx}} = \frac{{ - \sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}}\\\\
\text{Since} \qquad \qquad \qquad &\frac{{dy}}{{dx}} = \frac{{ - 1}}{2}, \text{we now get:}\\\\&\frac{{\sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}} = \frac{1}{2}\\\\ \Rightarrow  \qquad \qquad &\sin \left( {x + y} \right) = 1 \qquad \qquad  \dots\rm{(i)}\end{align}\]

But if  \(\sin \left( {x + y} \right)\) is 1, \(\cos \left( {x + y} \right)\)  must be 0, so that the equation of the curve reduces to \(y = \cos \left( {x + y} \right) = 0\).

Therefore, \(sin\,x=\text{ }1\)           (from (i))

\[\begin{align}&\Rightarrow \qquad x = \frac{{ - 3\pi }}{2},\frac{\pi }{2} \text{(in the given range for x)}\\ &\Rightarrow \qquad \text{There are two points on the curve at which the tangent drawn will have slope}\; \frac{{ - 1}}{2}; \text{namely} \left( {\frac{{ - 3\pi }}{2},0} \right)\,\,{\text{and}}\,\,\left( {\frac{\pi }{2},0} \right)\end{align}\]

\(\begin{align}\rm{\bf{Equations\;of\; tangent:}} \qquad y &= \frac{{ - 1}}{2}\left( {x - \frac{\pi }{2}} \right) \Rightarrow x + 2y = \frac{\pi }{2}\\\\ \text{and} \;\;y &= \frac{{ - 1}}{2}\left( {x + \frac{{3\pi }}{2}} \right) \Rightarrow x + 2y = \frac{{ - 3\pi }}{2}\end{align}\)

 

Example – 7

Find the point of intersection of the tangents drawn to the curve  \({x^2}y = 1 - y\) at the points where it is intersected by the curve \(xy=\text{ }1-y.\)

Solution: We first need to find out the points of intersection of the two curves before determining the equations of tangents at those points:

\[\begin{align}&{x^2}y = 1 - y\;\;\;\;\;\; \;\;\;\;\;\; ... \rm{(i)}\\\\&xy = 1 - y\;\;\;\;\;\;\; \;\;\;\;\;\;... \rm{(ii)}\\\\ \Rightarrow  \qquad &x\left( {1 - y} \right) = \left( {1 - y} \right)\\\\\Rightarrow \qquad &\left( {1 - x} \right)\left( {1 - y} \right) = 0\\\\ \Rightarrow \qquad & x = 1\,\,\,{\rm{or  }}y = 1\end{align}\]

The two points or \(\begin{align}\left( {1,\frac{1}{2}} \right)\end{align}\)  and  \(\left( {0,1} \right)\) \(\left( {{\rm{verify}}} \right)\)

Since we need the tangents to the first curve, we differentiate (i):

\[\begin{align}& 2xy+{{x}^{2}}\frac{dy}{dx}=\frac{-dy}{dx} \\  &  \\ 
 &\Rightarrow \text{ }\!\!~\!\!\text{ }\qquad \frac{dy}{dx}=\frac{-2xy}{1+{{x}^{2}}} \\ 
 &  \\ & {{m}_{{{T}_{1}}}}={{\left. \frac{dy}{dx} \right|}_{\left( 1,{}^{1}\!\!\diagup\!\!{}_{2}\; \right)}}=\frac{-1}{2}\,\,\text{and}\,\,{{m}_{{{T}_{2}}}}={{\left. \frac{dy}{dx} \right|}_{\left( 0,1 \right)}}=0 \\ 
\end{align}\]

\(\begin{align}\rm{\bf{Equations\;of\;tangents:}} \qquad &y - \frac{1}{2} = \frac{{ - 1}}{2}\left( {x - 1} \right) \Rightarrow x + 2y - 2 = 0 \qquad \dots\rm{(iii)}\\\\&y - 1 = 0\,\left( {x - 0} \right) \Rightarrow y = 1\qquad   \qquad \qquad \quad \; \dots\rm{(iv)} \end{align}\)

The point of intersection of (iii) and (iv) is clearly (0, 1).

 

Example – 8

Prove that the length intercepted by the co-ordinate axes on any tangent to the curve  \({x^{2/3}} + {y^{2/3}} = {c^{2/3}}\) is constant.

Solution: Differentiating the given curve w.r.t  x, we get:

\[\begin{align}&\frac{2}{3}{x^{ - 1/3}} + \frac{2}{3}{y^{ - 1/3}}\frac{{dy}}{{dx}} = 0\\\\
\Rightarrow  &\frac{{dy}}{{dx}} =  - {\left( {\frac{y}{x}} \right)^{1/3}}\end{align}\]

Let us take a point on this curve as \(\begin{align}\left( {t,{{\left( {{c^{2/3}} - {t^{2/3}}} \right)}^{3/2}}} \right)\end{align}\)

\[\begin{align}\Rightarrow  {m_T} = {\left. {\frac{{dy}}{{dx}}} \right|_{x = t}} &=  - {\left( {\frac{{{{\left( {{c^{2/3}} - {t^{2/3}}} \right)}^{3/2}}}}{t}} \right)^{1/3}}\\\\
 &=  - \frac{{{{\left( {{c^{2/3}} - {t^{2/3}}} \right)}^{1/2}}}}{{{t^{1/3}}}}\end{align}\]

Equation of tangent:      \(y - {\left( {{c^{2/3}} - {t^{2/3}}} \right)^{3/2}} = {m_T}\left( {x - t} \right)\)

           y – intercept:                       Put x = 0

\[\begin{align} \Rightarrow \qquad y &=  - t{m_T} + {\left( {{c^{2/3}} - {t^{2/3}}} \right)^{3/2}}\\\\
&= {t^{2/3}}{\left( {{c^{2/3}} - {t^{2/3}}} \right)^{1/2}} + {\left( {{c^{2/3}} - {t^{2/3}}} \right)^{3/2}}\\\\&= {\left( {{c^{2/3}} - {t^{2/3}}} \right)^{1/2}}\left\{ {{t^{2/3}} + {c^{2/3}} - {t^{2/3}}} \right\}\\\\&= {c^{2/3}}{\left( {{c^{2/3}} - {t^{2/3}}} \right)^{1/2}}\end{align}\]

x –intercept:                 Put y = 0

\[\begin{align}\Rightarrow x &= t - \frac{{{{\left( {{c^{2/3}} - {t^{2/3}}} \right)}^{3/2}}}}{{{m_T}}} \\\\&= t + \left( {{c^{2/3}} - {t^{2/3}}} \right){t^{1/3}}\\\\ &= {c^{2/3}}\,{t^{1/3}}\\\\{\text{Length intercepted between the axes }}&={(x-{\text{intercept)}^2}}{\rm{ }} + {\text{ }}{(y-{\text{intercept)}^2}}\\\\&= {c^{4/3}}{t^{2/3}} + {c^{4/3}}\left( {{c^{2/3}} - {t^{2/3}}} \right)\\\\&= {c^2}\end{align}\]

We see that the length is independent of the parameter t and is therefore a constant.