# Examples on Tangents and Chords of Parabolas Set 2

**Example – 34**

Find the locus of the point of intersection of tangents drawn at the extremities of a normal chord to the parabola \({y^2} = 4ax.\)

**Solution: ** Let \({t_1}\) be a point at which a normal is drawn to the parabola. This intersects the parabola again at the point

\[{t_2} = - {t_1} - \frac{2}{{{t_1}}}\qquad\qquad \qquad \dots \left( 1 \right)\]

Let the point of intersection of the tangents at \({t_1}\)and \({t_2}\) be \(P(h,k).\) From an earlier obtained result, we have

\[\begin{align}& \qquad \;\;h = a{t_1}{t_2}{\rm{ }}\;\;\;\;\;k = a({t_1} + {t_2})\\\\&\Rightarrow \quad h = - a{t_1}\left( {{t_1} + \frac{2}{{{t_1}}}} \right)\,\,\,\,\,k = a\left( {\frac{{ - 2}}{{{t_1}}}} \right) \qquad \qquad \left( {{\rm{Using }}\left( 1 \right)} \right)\\\\&\Rightarrow \quad h = - at_1^2 - 2a \;\;\;\; k = - \frac{{2a}}{{{t_1}}}\\\\&\Rightarrow \quad \frac{{h + 2a}}{{ - a}} = \frac{{4{a^2}}}{{{k^2}}}\qquad \qquad \qquad \qquad \qquad \qquad\;\;({\rm{eliminating}}\;{t_1})\\\\&\Rightarrow \quad {k^2}(h + 2a) + 4{a^3} = 0\end{align}\]

The required locus of *P* is

\[{y^2}(x + 2a) + 4{a^3} = 0\]

**Example - 35**

Find the polar of the pole \(P({x_1},{y_1})\) with respect to the parabola \({y^2}\) = 4ax.

**Solution: ** Recall that the polar is the locus of the point of intersection of the two tangents drawn at the extremities of a variable chord *QR* passing through *P*.

Let *M*(*h, k*) be a point on the required polar corresponding to *QR*.

The chord of contact of the tangents drawn from *M*, i.e. *QR*, has the equation

\[\begin{array}{l}T(h,k) = 0\\\\ky = 2a(x + h)\end{array}\]

If this is to pass through \(P({x_1},{y_1}),\) we have

\[k{y_1} = 2a({x_1} + h)\]

This is the required locus; in terms of (*x, y*), the locus (polar) is

\[\begin{align}&\qquad \;\; y{y_1} = 2a(x + {x_1})\\\\& \Rightarrow \quad \boxed {{T({x_1},{y_1}) = 0}} :{\rm{Equation\, of\, polar\, of }}P({x_1},{y_1})\end{align}\]

Observe that the polar of the focus will simply be the directrix.

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