Examples on Normals To Parabolas Set 2

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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

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


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 P on the parabola \({y^2} = 4ax\) meets its axis in A and the y-axis in B. Let O be the origin. The rectangle OACB is completed. Find the locus of C.

Q. 3

Prove that the normal to \({y^2} = 4ax\) at a non-zero point whose x and y coordinates are equal, subtends a right angle at the focus.

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 ?