Example – 7

The normals at three points A, B and C on a parabola intersect at O. F is the focus of the parabola. Prove that

\[FA\cdot FB \cdot FC = \frac{l}{4} \cdot F{O^2}\]

where l is the length of the latus-rectum.

Solution: Assume the parabola to be \({y^2} = 4ax,\) and the point O to be (h, k). F is the point (a, 0).

 

Any normal to is \({y^2}= 4ax\) is

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

and since this passes through O(h, k), we have

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

The feet of the three normal correspond to A, B and C. Thus, if \({m_1},\,{m_2}{\;\rm{and }}\;{\rm{ }}{m_3}\) and are the roots of (1), the co-ordinates of A, B and C are

\[(am_1^2,\, -2a{m_1}),\,(am_2^2,\, - 2a{m_2}){\;\rm{and}}\;{\rm{ }}(am_3^2,\, - 2a{m_3})\]

We have now,

\[FA = \sqrt {{{(a -am_1^2)}^2} + {{(2a{m_1})}^2}} \]

\[ = a(1 + m_1^2)\]

Similarly, \(FB = a(1 + m_2^2){\;\rm{ and}}\;{\rm{ }}FC = a(1 + m_3^2)\).

Thus,

\[\begin{align}&FA \cdot FB \cdot FC= {a^3}(1 + m_1^2)(1 + m_2^2)(1 + m_3^2)\\\\&= {a^3}\left( {1 +m_1^2 + m_2^2 + m_3^2 + m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 +{{({m_1}{m_2}{m_3})}^2}} \right) \qquad \quad \dots \left( 2 \right)\end{align}\]

From(1), we know the values of

\[\left.\begin{align}{m_1} + {m_2} +{m_3} = 0\\{m_1}{m_2} +{m_2}{m_3} + {m_3}{m_1} = \left( {\frac{{2a - h}}{a}} \right)\\{m_1}{m_2}{m_3} = -\frac{k}{a}\end{align}\right\} \qquad \quad \dots\left( 3 \right)\]

Our task is to express the relation (2) in terms of the known quantities given by(3). This can be done as follows :

\[\begin{align}&\qquad \qquad \qquad \qquad m_1^2+ m_2^2 + m_3^3 = {({m_1} + {m_2} + {m_3})^2} - 2({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})\\\\&m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 = {({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})^2} - 2{m_1}{m_2}{m_3}({m_1} + {m_2} + {m_3})\end{align}\]

Substituting the appropriate values gives

\[\begin{align}FA\cdot FB \cdot FC &= {a^3}\left\{ {1 + {{\left( {\frac{{2a - h}}{a}} \right)}^2}- 2\left( {\frac{{2a - h}}{a}} \right) + \frac{{{k^2}}}{{{a^2}}}} \right\}\\\\ &=a\{ {(h - a)^2} + {k^2}\} \end{align}\]

which evidently equals \(a \cdot F{O^2}.\) Since \(l = 4a,\) we get the desired result:

\[FA \cdot FB \cdot FC = \frac{l}{4} \cdot F{O^2}\]

Example - 8

Normals are drawn form the point P with slopes \({m_1},{\rm{ }}{m_2},{m_3}\) to the parabola \({y^2} = 4x.\) If the locus of P with \({m_1}{m_2} = \alpha \) is a part of the parabola itself,then find \(\alpha \) .

Solution: If we let P be the point (h, k), we have

\[{m^3} + (2 - h)m + k = 0 \qquad \quad \dots \left( 1 \right)\]

Thus,

\[\begin{align}&\qquad \;{m_1}{m_2}{m_3}= - k\\\\&\Rightarrow \quad {m_3} = - \frac{k}{\alpha }\left( {{\rm{since}}\;{m_1}{m_2} = \alpha } \right)\end{align}\]

Substituting this back into (1), we obtain

\[\begin{align}& \qquad \qquad - \frac{{{k^3}}}{{{\alpha ^3}}} - \frac{k}{\alpha }(2 - h) + k = 0\\\\&\Rightarrow \qquad \;\;{k^2} = {\alpha ^2}h - 2{\alpha ^2} + {\alpha ^3} \qquad  \qquad \dots \left( 2\right)\end{align}\]

Also, since P lies on the parabola itself, we have

\[{k^2} =4h...\left( 3 \right)\]

From(2) and (3), we have

\[\begin{array}{l}{\alpha^2} = 4{\rm{ }}\;and\;{\rm{ }}{\alpha ^3} - 2{\alpha ^2} = 0\\\\\Rightarrow \alpha = 2\end{array}\]

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