Parabola
12.0 Important relations
12.0 Important relations
1. In general, maximum three normals can be drawn from a point $P(h,k)$ to the parabola.
Proof: Let the equation of parabola be ${y^2} = 4ax$ and the equation of normal to the parabola in slope form is $$y = mx - 2am - a{m^3}$$
If it passes through point $P(h,k)$, then point satisfies the equation. We get $$k = mh - 2am - a{m^3}$$
$$a{m^3} + m\left( {2a - h} \right) + k = 0...(1)$$
Equation $(1)$ is a cubic equation in $m$, so it has three roots, say ${m_1}$, ${m_2}$ and ${m_3}$ such that
$${m_1} + {m_2} + {m_3} = 0...(2)$$
$${m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = \frac{{2a - h}}{a}...(3)$$ and $${m_1}{m_2}{m_3} = \frac{{ - k}}{a}...(4)$$
Hence for any given point $P(h,k)$, equation $(1)$ has three real or imaginary roots and corresponding to each of these three roots, we have one normal each passing through a point $P(h,k)$. Therefore, maximum three normals can be drawn from a point $P(h,k)$ to the parabola.
Points $A$, $B$ and $C$ in which three normals from a point $P(h,k)$ meet the parabola is called co-normal points.
2. If three distinct normals are drawn from a point $P(h,k)$ to the parabola, then the sum of the slopes of the normal is $0$.
Proof: As shown in equation $(2)$ in part $(a)$ i.e., ${m_1} + {m_2} + {m_3} = 0$. Since the coefficient of ${m^2}$ in the equation $(1)$ in part $(a)$ is $0$, sum of slopes is also $0$.
3. The algebraic sum of ordinates of the feets of three normals drawn to a parabola from a given point is $0$.
Proof: Let us assume the ordinates of points $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ in terms of slope be
$${y_1} = - 2a{m_1},{\text{ }}{y_2} = - 2a{m_2},{\text{ }}{y_3} = - 2a{m_3}$$
Therefore, the algebraic sum of these ordinates is $${y_1} + {y_2} + {y_3} = - 2a{m_1} - 2a{m_2} - 2a{m_3}$$ $$ = - 2a\left( {{m_1} + {m_2} + {m_3}} \right)$$ $$ = - 2a \times 0{\text{ [from part (b)]}}$$ $$=0$$
4. If three normals are drawn from a point $P(h,k)$ to the parabola are real then $h>2a$.
Proof: When normals are real, then all the three roots of equation $(1)$ in part $(a)$ are real i.e., $${m_1}^2 + {m_2}^2 + {m_3}^2 > 0$$
or, $${({m_1} + {m_2} + {m_3})^2} - 2\left( {{m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1}} \right) > 0$$
From equation $(2)$, we get ${m_1} + {m_2} + {m_3} = 0$
Therefore, $${0^2} - \frac{{2\left( {2a - h} \right)}}{a} > 0$$ or, $$h-2a>0$$ $$h>2a$$
5. Centroid of a triangle formed by three co-normal points always lies on the axis of parabola.
Proof: Let us assume the coordinates of the vertices of $\Delta ABC$ be $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$, then its centroid is $$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},0} \right)$$
From part $(c)$, ${y_1} + {y_2} + {y_3} = 0$. Hence the centroid lies on the $x-$axis $OX$, which is also the axis of the parabola.
Now, $$\frac{{{x_1} + {x_2} + {x_3}}}{3} = \frac{1}{3}\left( {a{m_1}^2 + a{m_2}^2 + a{m_3}^2} \right)$$ $$\frac{{{x_1} + {x_2} + {x_3}}}{3} = \frac{a}{3}\left( {{m_1}^2 + {m_2}^2 + {m_3}^2} \right)$$
$$\frac{{{x_1} + {x_2} + {x_3}}}{3} = \frac{a}{3}\left\{ {{{({m_1} + {m_2} + {m_3})}^2} - 2\left( {{m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1}} \right)} \right\}$$
From equations $(2)$ and $(3)$,
$$\frac{{{x_1} + {x_2} + {x_3}}}{3} = \frac{a}{3}\{ {(0)^2} - 2\left( {\frac{{2a - h}}{a}} \right)\} $$
Therefore, Centroid of $\Delta ABC$ is $(\frac{{2h - 4a}}{3},0)$.
