Normals to Parabolas

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Many properties of normals in parabolas are non-trivial and important enough to be discussed independently of tangents. This is what we do here. Once again, we use the parabola \({y^2} = 4ax\) for illustration purposes although the discussion can easily be generalised.

EQUATION OF NORMAL AT \(P({x_1},{y_1})\) : Let \(P({x_1},{y_1})\) be a point on the parabola; thus \(y_1^2 = 4a{x_1}\) .The slope of the tangent at P is

\[{m_T} = {\left. {\frac{{dy}}{{dx}}} \right|_{({x_1},{y_1})}} = \frac{{2a}}{{{y_1}}}\]

Thus, the slope of the normal is

\[{m_N} = \frac{{ - {y_1}}}{{2a}}\]

The equation of the normal at \(({x_1},{y_1})\) can now be written using point-slope form:

\[\boxed{\begin{align}{y - {y_1} = \frac{{ - {y_1}}}{{2a}}(x - {x_1})}\end{align}}\]

EQUATION OF NORMAL AT P \(({\bf{a}}{{\bf{t}}^2},{\bf{2at}})\): The point P is now given in parametric form. Using the equation of the normal derived above, we can write the equation in parametric form by using \(\left( {a{t^2},{\rm{ }}2at} \right)\) instead of \(({x_1},{y_1})\) :

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

Note that this normal has slope m = –t. Thus, the same equation can also be specified in terms of slope as described below.

EQUATION OF NORMAL WITH SLOPE m: Instead of t, we use –m in the equation above :

\[\begin{align}& \qquad \;\;y - mx = - 2am - a{m^3}\\\\&\Rightarrow \quad \boxed{{y = mx - 2am - a{m^3}}}\end{align}\]

Note that this is the normal at the point \(\left( {a{t^2},{\rm{ }}2at} \right)\) which in terms of m is \(\left( {a{m^2},{\rm{ }}-2am} \right).\)

The cubic equation in m hints that from a given point P, three normals can be drawn to the parabola. Let us try to prove this. Suppose that the point P is (h, k). Since the normal (s) of slope m passes through P, we have

\[\begin{align}& \qquad \;\;k = mh - 2am - a{m^3}\\&\Rightarrow \quad a{m^3} + (2a - h)m - k = 0 \qquad \quad \dots\left( 1 \right)\end{align}\]

This gives three values of m, say \({m_1}{\rm{ }},{\rm{ }}{m_2}{\rm{ }}\,and\,{\rm{ }}{m_3}\) and thus three corresponding normals. However, the roots \({m_1}{\rm{ }},{\rm{ }}{m_2}{\rm{ }}\,and\,{\rm{ }}{m_3}\) may not all be real. Two of them could be imaginary (one will always be real). Thus, depending on the coefficients in(1), we could either have one or three normals from P to the parabola. (There is also the possibility of two identical roots in which case only two normals will actually exist)

In case there are three normals, these will intersect the parabola at \((am_i^2,\, - 2a{m_i})\) for i = 1, 2 3. The sum of the ordinates of these points is

\[ - 2a({m_1} + {m_2} + {m_3})\]

which is 0 from (1).

Example – 26

What are the points on the parabola \({y^2} = 4ax\) from which three distinct normals can be drawn to the parabola?

Solution: Assume a point \(P(a{t^2},2at)\) from which three normal to the parabola can be drawn. We basically need to find the range of the variable t for which this is possible. The equation of an arbitrary normal to this parabola can be written as

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

If this passes through P, we have

\[\begin{align}& \qquad \;\; 2at = a{t^2}m - 2am - a{m^3}\\\\&\Rightarrow \quad {m^3} + (2 - {t^2})m + 2t = 0\\\\& \Rightarrow \quad (m + t)({m^2} - mt + 2) = 0\end{align}\]

One root for m is –t which actually gives the normal at P itself. This should have been expected because a normal to the parabola itself on any point can obviously always be drawn.

The other two roots for m are real and distinct if

\[{t^2} > {\rm{ }}8\]

This is the condition that the parameter t must satisfy if we are to be able to draw three real and distinct normals from P to the parabola.

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