Question

The length of the chord along the normal at $\left( {ct,\dfrac{c}{t}} \right)$ on the curve $xy = {c^2}$ is ?

Answer 1

Here in this question we have to find the length of the chord, we know the value of one point but we don’t know the value of another point. First by using the equation of a normal and considering the point be $\left( {c{t_1},\dfrac{c}{{{t_1}}}} \right)$ we determine the actual value of the other point and hence by using the distance between the two points we determine the value of the length of chord.

Complete step by step answer:
Now consider the equation of a curve $xy = {c^2}$.
This can be written as
$\Rightarrow y = \dfrac{{{c^2}}}{x}$------ (1)
On differentiating the equation (1) with respect to x.
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{c^2}}}{{{x^2}}}$ ------(2)
Now we have to find the derivative value at the point $\left( {ct,\dfrac{c}{t}} \right)$. Here the value of x is $ct$ and the value of $\dfrac{c}{t}$. On substituting values in the equation (2) we get
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{c^2}}}{{{c^2}{t^2}}}$
On simplifying we have
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{{t^2}}}$
As we know that the equation of normal is given by $\left( {y - {y_1}} \right) = \dfrac{{ - 1}}{{\dfrac{{dy}}{{dx}}}}\left( {x - {x_1}} \right)$ at point $\left( {ct,\dfrac{c}{t}} \right)$. So we have
$\Rightarrow \left( {y - \dfrac{c}{t}} \right) = \dfrac{{ - 1}}{{\dfrac{{ - 1}}{{{t^2}}}}}\left( {x - ct} \right)$
On simplifying we have
$\Rightarrow \left( {y - \dfrac{c}{t}} \right) = {t^2}\left( {x - ct} \right)$
Simplify the LHS term by taking LCM we have
$\Rightarrow \left( {\dfrac{{yt - c}}{t}} \right) = {t^2}\left( {x - ct} \right)$
On multiplying by t on both sides we have
$\Rightarrow \left( {\dfrac{{yt - c}}{t}} \right) \times t = {t^2}\left( {x - ct} \right) \times t$
On simplifying we have
$\Rightarrow yt - c = {t^3}\left( {x - ct} \right)$
In the RHS multiply the terms we have
$\Rightarrow yt - c = {t^3}x - c{t^4}$
On rearranging we have
$\Rightarrow yt - {t^3}x = c - c{t^4}$
On dividing by t throughout the equation we have
$\Rightarrow y - {t^2}x = \dfrac{c}{t} - c{t^3}$ ------(3)
This meets the curve at say, $\left( {c{t_1},\dfrac{c}{{{t_1}}}} \right)$
On substituting the values of x and y in the equation (3) we have
$\Rightarrow \dfrac{c}{{{t_1}}} - {t^2}\left( {c{t_1}} \right) = \dfrac{c}{t} - c{t^3}$
On simplifying we have
$\Rightarrow \dfrac{c}{{{t_1}}} - c{t_1}{t^2} = \dfrac{c}{t} - c{t^3}$
Take LCM on both the sides we have
$\Rightarrow \dfrac{{c - c{t_1}^2{t^2}}}{{{t_1}}} = \dfrac{{c - c{t^4}}}{t}$
On cross multiplying we have
$\Rightarrow t\left( {c - c{t_1}^2{t^2}} \right) = {t_1}\left( {c - c{t^4}} \right)$
On simplifying we have
$\Rightarrow tc - c{t_1}^2{t^3} = {t_1}c - c{t_1}{t^4}$
Take c as a common on both sides we have
$\Rightarrow c\left( {t - {t_1}^2{t^3}} \right) = c\left( {{t_1} - {t_1}{t^4}} \right)$
On cancelling c on both sides we have
$\Rightarrow \left( {t - {t_1}^2{t^3}} \right) = \left( {{t_1} - {t_1}{t^4}} \right)$
On rearranging this we can write it as
$\Rightarrow t - {t_1} = {t_1}^2{t^3} - {t_1}{t^4}$
Take ${t_1}{t^3}$ as common in RHS we have
$\Rightarrow t - {t_1} = - {t_1}{t^3}(t - {t_1})$
On dividing by $t - {t_1}$ throughout the equation
$\Rightarrow \dfrac{{t - {t_1}}}{{t - {t_1}}} = \dfrac{{ - {t_1}{t^3}(t - {t_1})}}{{t - {t_1}}}$
So we have
$\Rightarrow 1 = - {t_1}{t^3}$
$\Rightarrow {t_1} = - \dfrac{1}{{{t^3}}}$
This point $\left( {c{t_1},\dfrac{c}{{{t_1}}}} \right)$ can be written as $\left( {\dfrac{{ - c}}{{{t^3}}},\dfrac{c}{{ - \dfrac{1}{{{t^3}}}}}} \right) = \left( { - \dfrac{c}{{{t^3}}}, - c{t^3}} \right)$
Let we consider the end points of the chord be P and Q, the point is $P\left( {ct,\dfrac{c}{t}} \right)$ and $Q\left( { - \dfrac{c}{{{t^3}}}, - c{t^3}} \right)$
The length of the chord is determined by formula $PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$-----(4)
Here the value of ${x_2} = ct$, ${x_1} = \dfrac{{ - c}}{{{t^3}}}$, ${y_2} = \dfrac{c}{t}$ and ${y_1} = - c{t^3}$. on substituting these values in equation (4) we have
$\Rightarrow PQ = \sqrt {{{\left( {ct + \dfrac{c}{{{t^3}}}} \right)}^2} + {{\left( {\dfrac{c}{t} + c{t^3}} \right)}^2}}$
$\Rightarrow PQ = \sqrt {\left( {{c^2}{t^2} + \dfrac{{{c^2}}}{{{t^6}}} + 2ct.\dfrac{c}{{{t^3}}}} \right) + \left( {\dfrac{{{c^2}}}{{{t^2}}} + {c^2}{t^6} + 2\dfrac{c}{t}.c{t^3}} \right)}$
$\Rightarrow PQ = \sqrt {{c^2}{t^2} + \dfrac{{{c^2}}}{{{t^6}}} + \dfrac{{2{c^2}}}{{{t^2}}} + \dfrac{{{c^2}}}{{{t^2}}} + {c^2}{t^6} + 2{c^2}{t^2}}$
$\Rightarrow PQ = \sqrt {3{c^2}{t^2} + \dfrac{{{c^2}}}{{{t^6}}} + \dfrac{{3{c^2}}}{{{t^2}}} + {c^2}{t^6}}$
$\Rightarrow PQ = c\sqrt {3{t^2} + \dfrac{1}{{{t^6}}} + \dfrac{3}{{{t^2}}} + {t^6}}$
$\Rightarrow PQ = c\sqrt {{{\left( {{t^2} + \dfrac{1}{{{t^2}}}} \right)}^3}}$
$\Rightarrow PQ = c\sqrt {{{\left( {\dfrac{{{t^4} + 1}}{{{t^2}}}} \right)}^3}}$
$\Rightarrow PQ = \dfrac{{c{{\left( {{t^4} + 1} \right)}^{\dfrac{3}{2}}}}}{{{t^{\dfrac{6}{2}}}}}$
On simplifying we have
$\therefore PQ = \dfrac{c}{{{t^3}}}{\left( {{t^4} + 1} \right)^{\dfrac{3}{2}}}$

Therefore the length of the chord along the normal at $\left( {ct,\dfrac{c}{t}} \right)$ on the curve $xy = {c^2}$ is $\dfrac{c}{{{t^3}}}{\left( {{t^4} + 1} \right)^{\dfrac{3}{2}}}$.

Note: Usually when we have to determine the length of the chord we use the formula $2\sqrt {{r^2} - {d^2}}$, where r is the radius and d is the diameter of a circle. But here we don’t know the value of radius and diameter so we take the distance between the two pints formula and that will be the length of the chord.