# Examples on Normals to Parabolas Set 1

**Example - 27**

A normal is drawn to the parabola \({y^2} = 4ax\) at the point *t*. Find the other point at which this normal intersects the parabola.

**Solution: ** The equation of the normal at *t* is

\[y + tx = 2at + a{t^3} \qquad \dots\left( 1 \right)\]

If this intersects the parabola again at \({{t}_{1}}\text{ then}\;\;(at_{1}^{2},\text{2}a{{t}_{1}})\) must satisfy (1). Thus,

\[\begin{align} & \qquad \;\;2a{t_1} + at_1^2t = 2at + a{t^3}\\\\&\Rightarrow \quad at(t_1^2 - {t^2}) = 2a(t - {t_1})\\\\&\Rightarrow \quad t({t_1} + t) = - 2\\&\Rightarrow \quad \boxed{{{t_1} = - t - \frac{2}{t}}} \end{align}\]

This very frequently used result tells us how to find the other end point of the normal chord which is normal at a point *t* to \({y^2} = 4ax.\)

Sometimes, a question might be posed pertaining to normals. Instead of solving it entirely from scratch, we could use already known properties that we might have discussed earlier, say, in the section on tangents. This saves a lot of time in a subject like co-ordinate geometry.

For example, suppose *AB* is a focal chord of a parabola. What is the angle between the tangent at *A* and the normal at *B*?

We discussed earlier that tangents at extremities of any focal chord are perpendicular. Thus, a tangent at one end-point and a normal at the other must be parallel !

Thus, you can see that the skill you need to master for this subject is to remember certain well known and frequently used results and use them to your advantage as much as possible.

**Example – 28**

The normals at \({t_1}{\rm{ }}\;and\;{\rm{ }}{t_2}\) of the parabola \({y^2} = 4ax.\) meet at \({t_3} \). Prove that \({t_1}{t_2} = 2.\)

**Solution: ** We can very easily solve this question using the result of the last example. Since the normal drawn at \({t_1}\) intersects the parabola again in \({t_3} \), we have \({t_3} = - {t_1} - \begin{align}\frac{2}{{{t_1}}}\end{align}.\) Similarly, \({t_3} = - {t_2} - \begin{align}\frac{2}{{{t_2}}}\end{align}.\) Comparing the two gives \({t_1}{t_2} = 2.\) But let us solve it without using this result. The equation of the normals at \({t_1}{\rm{ }}\,and\,{\rm{ }}{t_2}\) are

\[\begin{align}&y + {t_1}x = 2a{t_1} + at_1^3 \qquad \dots \left( 1 \right)\\\\&y + {t_2}x = 2a{t_2} + at_2^3 \qquad \dots \left( 2 \right)\end{align}\]

Both of these must be satisfied by the point \({t_3} \) , i.e., by \((at_3^2,\;2a{t_3}).\) Thus,

\[\begin{align}&2a{t_3} + a{t_1}t_3^2 = 2a{t_1} + at_1^3 \qquad \dots \left( 3 \right)\\\\&2a{t_3} + a{t_2}t_3^2 = 2a{t_2} + at_2^3 \qquad \dots \left( 4 \right)\end{align}\]

Subtracting(4) from (3), we have

\[\begin{align}& \qquad \;\;at_3^3({t_1} - {t_2}) = 2a({t_1} - {t_2}) + a(t_1^3 - t_2^3)\\\\& \Rightarrow \quad t_3^2 = 2 + t_1^2 + t_2^2 + {t_1}{t_2} & \dots \left( 5 \right)\end{align}\]

Using (5) in (3), we finally have

\[\begin{align}& \qquad \;\;4{a^2}(2 + t_1^2 + t_2^2 + {t_1}{t_2}) = {(a{t_1}t_2^2 + at_1^2{t_2})^2}\\\\&\Rightarrow \quad 8 + 4\{ {({t_1} + {t_2})^2} - {t_1}{t_2}\} = t_1^2t_2^2{({t_1} + {t_2})^2}\\\\&\Rightarrow \quad ({t_1}{t_2} - 2)\{ {({t_1} + {t_2})^2}({t_1}{t_2} + 2) + 4\} = 0\end{align}\]

which gives \({t_1}{t_2} = 2,\) as required.

Four important properties that tangents in any parabola satisfy have been listed. Since a normal at a point is perpendicular to the tangent at that point, we can use the properties of tangents to deduce the corresponding properties for normals.

\(*~\left( A \right)\) |
Tangent at any point bisects the angle \(\theta \) between the focal chord through that point and the perpendicular to the directrix from that point |
\( \Rightarrow \) |
Normal at any point bisects the |

\[*~\left( B \right)\] |
The tangent at one extremity of any focal chord parabola is perpendicular other to the normal at the other extremity. |
\( \Rightarrow \) |
The tangent at one extremity of any focal chord of a parabola is parallel to the normal at the extremity (We’ve already stated this earlier) |

Note that from property (A) above, we can also deduce that the normal at any point of a parabola is equally inclined to the focal chord through that point and the axis of the parabola. You are urged to prove this independently as an exercise.

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