# Examples on Tangents and Normals Set 2

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## Tangents and normals

### Example – 4

Tangents are drawn to the ellipse $${x^2} + 2{y^2} = 2.$$  Find the locus of the mid-point of the intercept made by the tangent between the co-ordinate axes.

Solution: To determine the required locus, we first write down the equation of an arbitrary tangent to the given ellipse:

$\frac{{{x^2}}}{2} + {y^2} = 1$

A general point on this ellipse can be taken as $$\left( {\sqrt 2 \cos \theta ,\sin \theta } \right)$$. Now we write the equation of the tangent at this point by first differentiating the equation of the ellipse:

\begin{align}&x + 2y\frac{{dy}}{{dx}} = 0\\\\ \Rightarrow \qquad \qquad &\frac{{dy}}{{dx}} = \frac{{ - x}}{{2y}}\\\\ \Rightarrow \qquad \qquad &{m_T}\left( {{\rm{at}}\left( {\sqrt 2 \cos \theta ,\sin \theta } \right)} \right) = \frac{{ - \sqrt 2 \cos \theta }}{{2\sin \theta }} = \frac{{ - \cot \theta }}{{\sqrt 2 }} \end{align}

\begin{align} &\mathbf{Equation}\text{ }\mathbf{of}\text{ }\mathbf{tangent}: \qquad y-\sin \theta =\frac{-\cot \theta }{\sqrt{2}}\left( x-\sqrt{2}\cos \theta \right) \\\\ &\quad\qquad \qquad \qquad \qquad \Rightarrow \qquad x\cos \theta +\sqrt{2}y\sin \theta =\sqrt{2} \\\\&x-\mathbf{intercept}:\qquad \text{Put} \qquad y=0\qquad\Rightarrow \ \ \ \ \ x=\sqrt{2}\sec \theta \\\\& \qquad \qquad \qquad \qquad \qquad \Rightarrow \qquad \text {The tangent intersects the \(x -axis at}\;P\left( \sqrt{2}\sec \theta ,0 \right) \\\\& y-\mathbf{intercept}:\qquad \text{Put }\qquad x\text{ }=\text{ }0\qquad\Rightarrow \qquad y=\;cosec\,\theta \\\\& \qquad \qquad \qquad \qquad \qquad \quad \Rightarrow \qquad  \text{The tangent intersects the y-axis at}\,\text{Q}\,(0,\,\,cosec\,\theta \text{ }) \end{align}\)

We require the locus of R, the mid-point of PQ. Let its co–ordinates be (h, k). Therefore,

\begin{align}h = \frac{{\sqrt 2 \sec \theta }}{2} \Rightarrow \cos \theta = \frac{1}{{\sqrt 2 h}} \qquad \qquad \dots \text{(i)} \end{align}

\begin{align}k = \frac{{\cos {\rm{ec}}\theta }}{2} \Rightarrow \sin \theta = \frac{1}{{2k}} \qquad \qquad \;\;\;\;\;\;\dots \text{(ii)}\end{align}

Squaring and adding (i) and (ii), we get:

$\frac{1}{{2{h^2}}} + \frac{1}{{4{k^2}}} = 1$

Therefore, the locus of R is:

$\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$

### Example – 5

Find the point on the ellipse $$4{x^2} + 9{y^2} = 1$$ at which the tangent is parallel to the line $$8x=\text{ }9y.$$

Solution: The ellipse can be rewritten as:

$\frac{{{x}^{2}}}{{}^{1}\!\!\diagup\!\!{}_{4}\;}+\frac{{{y}^{2}}}{{}^{1}\!\!\diagup\!\!{}_{9}\;}=1$

Let a general point on this ellipse be \begin{align}\left( {\frac{1}{2}\cos \theta ,\,\,\frac{1}{3}\sin \theta } \right).\end{align}

Differentiating the equation of the given ellipse, we get:

\begin{align}&8x + 18y\frac{{dy}}{{dx}} = 0\\\\ \Rightarrow \qquad &\frac{{dy}}{{dx}} = \frac{{ - 4x}}{{9y}}\\\\\text{At} \left( {\frac{1}{2}\cos \theta ,\,\,\frac{1}{3}\sin \theta } \right),\,\,\,{m_T} &= \frac{{ - 4 \times\begin{align} \frac{1}{2}\end{align}\cos \theta }}{{9 \times \begin{align}\frac{1}{3}\end{align}\sin \theta }} = \frac{{ - 2\cot \theta }}{3}\end{align}

For the tangent to be parallel to $$8x = 9y,\,\,\,{m_T}$$  must be equal to the slope of this line. Hence:

\begin{align}&\frac{{ - 2\cot \theta }}{3} = \frac{8}{9}\\\\\Rightarrow \qquad &\cot \theta = \frac{{ - 4}}{3}\\\\\Rightarrow \qquad &\sin \theta = \frac{3}{5},\cos \theta = \frac{{ - 4}}{5}\,\,\,{\rm{or}}\,\,\,\sin \theta = \frac{{ - 3}}{5},\,\,\cos \theta = \frac{4}{5}\\\\\text{Hence, the required point is} &\left( {\frac{1}{2}\cos \theta ,\,\,\frac{1}{3}\sin \theta } \right) \rm{ or:}\\\\&\left( {\frac{{ - 2}}{5},\frac{1}{5}} \right)\,\,{\rm{and}}\,\,\left( {\frac{2}{5},\frac{{ - 1}}{5}} \right)\end{align}

Applications of Derivatives
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