Question

Prove that the sum of intercepts which the normal cut-off from the axis is
\[2(h + a)\].

Answer 1

Hint:The given question asks us to prove that the sum of intercepts cutoff from the axis by the normal of a parabola is \[2(h + a)\]. The equation of a parabola is $y = 4ax$. The term $h$ here denotes the $x$ coordinate of the vertex of the parabola. We will first write the equation of normal of the parabola in the slope form then solve to find the sum of intercepts from the axis.

Complete step by step answer:
The equation of the slope of a parabola in normal form is given by,
$y = mx - 2am - a{m^3}$
It passes through $(h,k)$as it is the vertex of the parabola, so we can write,
$k = mh - 2am - a{m^3}$
$\Rightarrow a{m^3} + (2a - h)m + k = 0$
since the equation is cubic in $m$the sum of the roots is given by,
${m_1} + {m_2} + {m_3} = \dfrac{{ - 0}}{a} = 0$
The sum of roots multiplied two at a time will be given by
${m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = \dfrac{{2a - h}}{a}$
The product of the roots will be given by,
${m_1}{m_2}{m_3} = \dfrac{{ - k}}{a}$
These three equations are derived from the properties of the cubic equations.

Now we have got three relations between the three values of ${m_1},{m_2}\,and\,{m_3}$. In the original equation of the normal we will now find the intercepts by putting \[y = 0\] we get,
$x = 2a + a{m^2}$
Since there are three intercepts there sum $S$ will be,
$S = 2a + a{m_1}^2 + 2a + a{m_2}^2 + 2a + a{m_3}^2$
Solving it we get,
$S = 6a + a({m_1}^2 + {m_2}^2 + {m_3}^2)$
We can write
${m_1}^2 + {m_2}^2 + {m_3}^2$ $ = $$[{({m_1} + {m_2} + {m_3})^2} - 2({m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1})]$
Thus $S$ can be written as,
$S = 6a + a\{ {({m_1} + {m_2} + {m_3})^2} - 2({m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1})\} $
$\Rightarrow S = 6a + a\left\{ {{{(0)}^2} - 2\left( {\dfrac{{2a - h}}{a}} \right)} \right\}$
This is given by using the three equations we found by using the property of the cubic equation.
$S = 6a - 4a + 2h$
Which upon solving gives,
$\therefore S = 2a + h$
Hence proved.

Note:The normal of a curve is a line which is perpendicular to the tangent(straight line touching the curve at exactly one point) of that curve. The normal of a conic can be found out by finding the point which touches the tangent and then finding the slope of the normal using the formula of product of slopes for perpendicular lines which is: ${m_1}{m_2} = - 1$. The equation will then be found out by using the point slope form of the line.