# Tangents to Parabolas

In this section, we discuss tangents to parabolas and various properties that they will satisfy. As always, \({y^2} = 4ax\) is the parabola used for illustration but the discussion can obviously be generalised.

**TANGENT AT \(P\left( {{x_1}{y_1}} \right)\) : **Suppose a point \(P\left( {{x_1}{y_1}} \right)\) is given on the parabola, and we need to find the equation of the tangent at *P*. The slope of the tangent can easily be evaluated by differentiating the equation of the parabola :

\[\begin{align}&\qquad \quad \;\;{y^2} = 4ax\\\\&\Rightarrow \qquad 2y\frac{{dy}}{{dx}} = 4a\\\\&\Rightarrow \qquad \frac{{dy}}{{dx}} = \frac{{2a}}{y}\\\\&\Rightarrow \qquad {\left. {\frac{{dy}}{{dx}}} \right|_{{\rm{at }}P({x_1},{y_1})}} = \frac{{2a}}{{{y_1}}}\end{align}\]

Thus, the equation of the tangent is

\[\begin{align}&\qquad \;\;y - {y_1} = \frac{{2a}}{{{y_1}}}(x - {x_1})\\\\&\Rightarrow \quad y{y_1} = 2ax - 2a{x_1} + y_1^2\\\\& \Rightarrow \quad y{y_1} = 2ax - 2a{x_1} + 4ax_1^2 \qquad (\because \; y_1^2 = 4a{x_1})\\\\&\Rightarrow \quad \boxed{{y{y_1} = 2a(x + {x_1})}}\qquad \qquad \qquad \dots(1)\end{align}\]

This equation is sometimes written concisely as \(T({x_1},{y_1}) = 0\) where the interpretation of *T* is understood to be as in (1)

**TANGENT AT **** \({\bf{P}}({\bf{a}}{{\bf{t}}^2},{\rm{ }}{\bf{2at}})\): ** A lot many times, the point P is specified in its parametric form and we need to find the equation of the tangent in terms of the parameter *t*. This can be done by replacing \(\left( {{x_1},{\rm{ }}{y_1}} \right)\) with \(\left( {a{t^2},{\rm{ }}2at} \right)\) in the equation of the tangent just obtained :

\[\begin{align}&\qquad \;\;y{y_1} = 2a(x + {x_1})\\\\&\Rightarrow \quad 2aty = 2a(x + a{t^2})\\\\&\Rightarrow \quad \boxed{{ty = x + a{t^2}}}\end{align}\]

This form of the tangent is very widely used and it is best for you to commit it to memory.

**TANGENT OF SLOPE m: ** Suppose that we are required to find the equation of the tangent with slope

*m*.We did so in the last section when we found the intersection of the line \(y = mx + c\) with the parabola and obtained the condition for tangency to be

\[\fbox{\({c = \frac{a}{m}}\)}\]

The equation of the tangent thus obtained was

\[\fbox{\({y = mx + \frac{a}{m}}\)}\]

This line is a tangent to the parabola for all non-zero values of *m*.

The same equation can be obtained using a derivatives approach also. You are urged to do so as an exercise.

**Example – 16**

Show that the tangents at the extremities of any focal chord of a parabola intersect at right angles at the directrix.

**Solution:** Let the extremities of the focal chord be \({t_1}\,and\,{\rm{ }}{t_2}\) so that \({t_1}{t_2} = {\rm{ }}-1\). The equations to the tangents at \({t_1}\,and\,{\rm{ }}{t_2}\) are

\[\begin{align}&{t_1}y = x + at_1^2\\&{t_2}y = x + at_2^2\end{align}\]

The intersection point can easily be evaluated by solving these two equations to be \((a{t_1}{t_2},a({t_1} + {t_2})) = ( - a,a({t_1} + {t_2}))\) since \({t_1}{t_2} = - 1.\) Since the *x*-coordinate of the point of intersection is –a, it lies on the directrix \(x + a = 0.\) Also, note that the slopes of the two tangents are \(\begin{align}{{m}_{1}}=\frac{1}{{{t}_{1}}}\,\rm{ }and\,\text{ }{{m}_{2}}=\frac{1}{{{t}_{2}}}\text{ }so\text{ }that~\text{ }{{m}_{1}}{{m}_{2}}=\frac{1}{{{t}_{1}}{{t}_{2}}}=-1.\end{align}\) Thus, the tangents intersect at right angles on the directrix.

Note that *T* will also be the point at which the circle drawn on *PQ* as diameter touches the directrix, since \(\angle PTQ\) is 90° which then becomes an angle in the semi circle.

**POINT OF INTERSECTION OF TANGENTS AT t _{1} and t_{2}. **: There’s a particular relation we obtained while solving the last example which is important enough to be highlighted explicitly. If tangents are drawn to \({y^2} = 4ax\;\) at \({t_1}{\text{ and}}\;{t_2}\), their point of intersection will be

\[\fbox{\({a{t_1}{t_2},\,a({t_1} + {t_2})}\)}\]

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