# Examples on Tangents and Normals Set 4

**Example – 9**

Find the angle of intersection between \({y^2} = 4x\,\,{\rm{and}}\,\,{x^2} = 4y\).

**Solution: **

There are two points of intersection, which can be obtained by simultaneously solving the equations for the two curves.

\[\begin{align}&{y^2} = 4x\,\,{\rm{and}}\,\,{x^2} = 4y\\\\

\Rightarrow \qquad \quad &{y^4} = 16{x^2} = 64y\\\\

\Rightarrow \qquad \quad &y\left( {{y^3} - 64} \right) = 0\\\\\Rightarrow \qquad \quad &y = 0,\,4\\\\ \Rightarrow \qquad \quad &\!\!\!\text{The points of intersection are (0, 0) and (4, 4). Let} \; m_{T_1}\; \rm{and}\; m_{T_2} \\&\!\!\text {represent the slopes of tangents to}\;x^2\,=\,4y\; \rm{and}\; y^2 = 4x \;\text {respectively.}\end{align}\]

\( \begin{align}\rm{\bf{At (0, 0):}} \qquad \qquad \qquad &{m_{{T_1}}} = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = \frac{x}{2} = 0\\\\&{m_{{T_2}}} = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = \frac{2}{y} = \infty\\\\\Rightarrow \qquad &\text{The angle between these two tangents is obviously 90° which is visually}\\&\text{clear from Fig – 6.}\end{align}\)

\(\begin{align}\rm{\bf{At (4, 4):}}\qquad \qquad \qquad &{m_{{T_1}}}{\left. { = \frac{{dy}}{{dx}}} \right|_{x = 4}} = \frac{x}{2} = 2\\\\&{m_{{T_2}}}{\left. { = \frac{{dy}}{{dx}}} \right|_{x = 4}} = \frac{2}{y} = \frac{1}{2}\end{align}\)

The angle of intersection is:

\[\theta ={{\tan }^{-1}}\left( \frac{2-{}^{1}\!\!\diagup\!\!{}_{2}\;}{1+2\times {}^{1}\!\!\diagup\!\!{}_{2}\;} \right)={{\tan }^{-1}}\left( \frac{3}{4} \right)\]

Therefore, the two curves intersect in two points, once at \({{90}^{\circ }}\) and once at \(\begin{align}{\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\end{align}\)

**Example – 10**

Find the shortest distance between two points, one of which lies on the curve \({y^2} = 4ax,\) and the other on the circle \({x^2} + {y^2} - 24ay + 128{a^2} = 0\)

**Solution: **Notice that the circle’s equation can be written equivalently as

\[{\left( {x - 0} \right)^2} + {\left( {y - 12a} \right)^2} = {\left( {4a} \right)^2}\]

so that its centre is (0, 12*a*) and radius is 4*a*.

Let *MN* represent the shortest distance between the circle and the ellipse. Since the point *N* on the ellipse is nearest to the circle, it will also be nearest to the centre of the circle *O*, from amongst all the other points on the ellipse. Hence, to determine *MN*, we may equivalently find the shortest distance between the circle’s centre and any point on the ellipse.

Now, from Fig-8’s geometry, notice a very important fact. The tangent drawn at *M* must be perpendicular to *ON*, or equivalently, ** ON must be a normal to the ellipse**. Only then will

*N*be the closest point on the ellipse from

*O*. (Convince yourself that this should be true)

We take an arbitrary point on the parabola as \((at{^2},{\rm{ }}2at)\). We will write the normal to the parabola at this point and make this normal pass through the point *O*.

\[\begin{align}&{y^2} = 4ax\\\\ \Rightarrow \qquad &2y\frac{{dy}}{{dx}} = 4a\\\\\Rightarrow \qquad &\frac{{dy}}{{dx}} = \frac{{2a}}{y}\\\\&{m_N}\left( {at\left( {a{t^2},2at} \right)} \right) = {\left. {\frac{{ - dx}}{{dy}}} \right|_{\left( {a{t^2}, \, 2at} \right)}} = - t \end{align}\]

\(\begin{align}\rm{\bf{Equation \;\;of\;\; normal:}}\qquad \qquad \qquad &y - 2at = - t\left( {x - a{t^2}} \right)\\\Rightarrow \qquad &tx + y = 2at + a{t^3} \qquad \qquad \dots \rm{(i)}\end{align}\)

So that this normal passes through *O*, the co-ordinates of *O*(0, 12*a*) must satisfy (i)

\[\begin{align} \Rightarrow \qquad &12a = 2at + a{t^3}\\\\ \Rightarrow \qquad &2t + {t^3} = 12\\\\ \Rightarrow \qquad &t = 2\;\;\;\;\;\;\;\; \rm{(verify)} \end{align}\]

We therefore get the co-ordinates of *N* as (4*a*, 4*a*).

\(\begin{align}\text{Hence,} \qquad \qquad \qquad &ON = \sqrt {{{\left( {4a - 0} \right)}^2} + {{\left( {4a - 12a} \right)}^2}} = \sqrt {80{a^2}} = 4\sqrt 5 a\\\\\Rightarrow \qquad &MN = 4\sqrt 5 a - 4a = \,\,\,4\left( {\sqrt 5 - 1} \right)a \end{align}\)

**TRY YOURSELF - I**

**Q. 1 ** Find the equation of the tangent to the curve \(\begin{align}{\left( {\frac{x}{a}} \right)^n} + {\left( {\frac{y}{b}} \right)^n} = 20\end{align}\) at the point (*a*, *b*).

**Q. 2 ** If *l*_{1} and *l*_{2} are the lengths of the perpendiculars drawn from the origin on the gangent and normal to the curve \(\begin{align}{x^{2/3}} + {y^{2/3}} = {a^{2/3}}\end{align}\) respectively at an arbitrary point, find the value of \(\begin{align}4l_1^2 + l_2^2\end{align}\) in terms of *a*.

**Q. 3 ** Find the sum of the intercepts made on the coordinate axes by any tangent to the curve \(\sqrt x + \sqrt y = \sqrt a \)

**Q. 4 ** Find the angle at which the curves \(2{y^2} = {x^3}\;{\rm{and}}\;{y^2} = 32x\) intersect at the origin.

**Q. 5 ** If the tangent at \(\left( {{x_1},{y_1}} \right)\) to \({x^3} + {y^3} = {a^3}\) meets the curve again at \(\left( {{x_2},{y_2}} \right),\) find the value of \(\begin{align}\frac{{{x^2}}}{{{x_1}}} + \frac{{{y_2}}}{{{y_1}}}.\end{align}\)

**Q. 6 ** If \(ax + by = 1\) is a normal to the parabola \({y^2} = 4cx\), find the value of \(\begin{align}\frac{{{b^2} - c{a^3}}}{{a{b^3}c}}.\end{align}\)

**Q. 7 ** Do the curves \(\begin{align}xy = 4\;{\rm{and}}\;{x^2} + {y^2} = 8\end{align}\) touch each other or intersect?

**Q. 8 ** Find the equations of tangents drawn from (1, 2) to be curve \(\begin{align}{y^2} - 2{x^3} - 4y + 8 = 0\end{align}\)

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