Question

Prove that:
The tangent at the extremity of a diameter of a parabola is parallel to the system of chords it bisects.

Answer 1

Hint: In order to prove that the tangent is parallel to the system of the chords, we consider two points where the diameter of the parabola touches either ends. We find out the slopes of lines from points on either end to the midpoint. Now we find the tangent from one point and normal from the other point and compute their respective slopes. They are parallel if their slopes are equal.

Complete step by step answer:

Let us consider a parabola through which its line of diameter passes through its center, which touches the parabola at points P and Q respectively. They are from PS and QS at the center.

It is as shown in the figure as follows:

Now we know a point on a parabola is of the form, $\left( {{\text{a}}{{\text{t}}^2}{\text{, 2at}}} \right)$. Therefore let the coordinates of point P and Q be $\left( {{\text{a}}{{\text{t}}_1}^2{\text{, 2a}}{{\text{t}}_1}} \right)$and $\left( {{\text{a}}{{\text{t}}_2}^2{\text{, 2a}}{{\text{t}}_2}} \right)$respectively.

We know the general equation of a parabola is ${{\text{y}}^2} = 4{\text{ax}}$

Now from the figure the slopes of the lines PS and QS should be equal as they are a part of the same line.

We know the formula to determine the slope of a line passing through a point is given by, $\dfrac{{{{\text{y}}_2} - {{\text{y}}_1}}}{{{{\text{x}}_2} - {{\text{x}}_1}}}$

Now slope of PS = slope of QS

$\Rightarrow \dfrac{{{\text{2a}}{{\text{t}}_1}}}{{{\text{a}}{{\text{t}}_1}^2 - {\text{a}}}}$ = $\dfrac{{{\text{2a}}{{\text{t}}_2}}}{{{\text{a}}{{\text{t}}_2}^2 - {\text{a}}}}$

$\Rightarrow \dfrac{{{{\text{t}}_1}}}{{{{\text{t}}_1}^2 - 1}} = \dfrac{{{{\text{t}}_2}}}{{{{\text{t}}_2}^2 - 1}}$

$\Rightarrow {{\text{t}}_1}{{\text{t}}_2} = - 1{\text{ - - - }}\left( 1 \right)$

Now let us consider the extremity at P, the equation of a tangent passing through point P is given by

${\text{y = x + at}}_1^2$

We know the slope of a line passing through a point is given by, ${\text{m = }}\dfrac{{\text{y}}}{{\text{x}}}$

Hence the slope of this tangent passing through the point P is given by, ${\text{m = }}\dfrac{1}{{{{\text{t}}_1}}}$

Now let us consider a normal passing through the point Q which is given by,

${\text{y + }}{{\text{t}}_2}{\text{x = 2a}}{{\text{t}}_2}{\text{ + at}}_2^3$

The slope of this line is, ${\text{y = }}{{\text{-t}}_{\text{2}}}{\text{x }} \Rightarrow {\text{m = - }}{{\text{t}}_2}$

Now from the equation (1)

${\text{ - }}{{\text{t}}_2}$ can be expressed as $\dfrac{1}{{{{\text{t}}_1}}}$ which is nothing but the slope of tangent from point P. Therefore the slopes of both lines are equal.

Hence the tangent at the extremity of a diameter of a parabola is parallel to the system of chords it bisects.

Note: In order to solve this type of problems the key is to know the concept and formulae involved in parabola and straight lines. The general equation of a parabola, point on a parabola, slope of a line given a point, equation of tangent and equation of normal have to be known.

We must know that two lines are always parallel if their slopes are equal and perpendicular if the product of their slopes is equal to -1.