# Examples on Normals To Parabolas Set 2

**Example - 29**

Find the locus of the point of intersection of normals drawn to the parabola \({y^2} = 4ax\) at the extremities of a chord which subtends a right angle at the vertex of the parabola.

**Solution: ** Let \(A(at_1^2,\,2a{t_1}){\rm{ }}\;and\;{\rm{ }}B(at_2^2,\,2a{t_2})\) be the extremities of a chord which subtends a right angle at the vertex (0, 0);

Since \(OA \bot OB,\) we have

\[\begin{align}& \qquad \underbrace {\left( {\frac{{2a{t_1} - 0}}{{at_1^2 - 0}}} \right)}_{{\rm{slope\;of \;}}OA} \times \underbrace {\left( {\frac{{2a{t_2} - 0}}{{at_2^2 - 0}}} \right)}_{{\rm{slope\;of\;}}OB} = - 1\\&\Rightarrow \qquad {t_1}{t_2} = - 4\end{align}\]

The equation to \({N_1}\;and\;{\rm{ }}{N_2}\) can be written using the standard form of a normal at a point *t* :

\[\begin{align}&{N_1}:y + {t_1}x = 2a{t_1} + at_1^3\\\\&{N_2}:y + {t_2}x = 2a{t_2} + at_2^3\end{align}\]

Let the intersection of \({N_1}\,and\,{\rm{ }}{N_2}\) be the point *C*(*h, k*). The co-ordinates of *C* can be evaluated by solving the equations of \({N_1}and{\rm{ }}{N_2}\) simultaneously :

\[\begin{align}h &= 2a + a(t_1^2 + t_2^2 + {t_1}{t_2})\\\\&= - 2a + a(t_1^2 + t_2^2)\\\\\rm{and} \qquad {\rm{ }}k &= - a{t_1}{t_2}({t_1} + {t_2})\\\\&= 4a({t_1} + {t_2})\end{align}\]

We thus have, by eliminating \({t_1}\,and\,{\rm{ }}{t_2}\) , a relation in *h* and *k*:

\[\begin{align}\frac{{{k^2}}}{{16{a^2}}} &= \frac{{h + 2a}}{a} - 8\\\\\Rightarrow \qquad {k^2} &= 16a(h - 6a)\end{align}\]

Using (*x, y*) instead of (*h, k*), the required locus is

\[{y^2} = 16a(x - 6a)\]

**Example – 30**

Find the locus of the point of intersection of the three normals to the parabola \({y^2} = 4ax,\) two of which are inclined at right angles to each other.

**Solution: ** Let *P*(*h, k*) be the point whose locus we wish to determine. Any normal to the given parabola can be written as

\[y = mx - 2am - a{m^3}\]

If this passes through *P*(*h, k*), we have

\[\;\;\quad k = mh - 2am - a{m^3} \qquad \dots \left( 1 \right)\]

Let \({m_1},{\rm{ }}{m_2}{\rm{ }}\;and\;{\rm{ }}{m_3}\) be the three roots of this cubic. It is given that two of the normals are perpendicular, implying that the product of two of these three slopes, say \({m_1}\;and\;{\rm{ }}{m_2}\) is –1, i.e.\({m_1}{m_2}{\rm{ }} = {\rm{ }}-1.\)

From (1), we have

\[\begin{align}{m_1}{m_2}{m_3} &= \frac{{ - k}}{a}\\\\\Rightarrow \qquad {m_3} &= \frac{k}{a}\end{align}\]

Substituting this value of \({m_3}\) back in (1), we obtain a relation between *h* and *k* :

\[\begin{align}& \qquad \;\; k = \frac{{hk}}{a} - 2k - \frac{{a{k^3}}}{{{a^3}}}\\\\&\Rightarrow \quad k({k^2} + (3a - h)a) = 0\end{align}\]

Using (*x, y*) instead of (*h, k*) the required locus is

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

**TRY YOURSELF - III **

Q. 1 |
Find the equation of the normal to \({y^2} = 4x\) which is perpendicular to \(2x + 6y + 5 = 0\) . |

Q. 2 |
The normal at any point |

Q. 3 |
Prove that the normal to \({y^2} = 4ax\) at a non-zero point whose |

Q. 4 |
Prove that the normal at any point to a parabola is equally inclined to the focal chord passing through that point and the axis of the parabola. |

Q. 5 |
Have you ever heard of parabolic mirrors being very effective in concentrating incident light energy at a particular a point ? Can you think of a reason ? What would that particular point be ? |

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