Question

Length of the focal chord of the parabola ${y^2} = 4ax$ at a distance p from the vertex is
$\left( a \right)\dfrac{{2{a^2}}}{p}$
$\left( b \right)\dfrac{{{a^2}}}{{{p^2}}}$
$\left( c \right)\dfrac{{4{a^3}}}{{{p^2}}}$
$\left( d \right)\dfrac{{{p^2}}}{a}$

Answer 1

Hint: In this particular type of question use the concept that the parametric coordinates of any point on the parabola ${y^2} = 4ax$ is given as $\left( {a{t^2},2at} \right)$ and use the concept that focal chord of the parabola is always passing through the focus of the parabola (a, 0) so use these concepts to reach the solution of the question.

Complete step-by-step solution:

Let the focal chord of the parabola ${y^2} = 4ax$ be AA’ as shown in the above figure, passing from the focus (a, 0) the length of this chord from the vertex is at distance p as shown in the above figure.

As we all know that the parametric coordinates of any point on the parabola ${y^2} = 4ax$ is given as $\left( {a{t^2},2at} \right)$.

So let A = $\left( {a{t^2},2at} \right)$ and A’ = $\left( {at_1^2,2a{t_1}} \right)$

Now the focal chord is passing from the focus F (a, 0)

So the slope of AF = slope of FA’

$ \Rightarrow \dfrac{{0 - 2at}}{{a - a{t^2}}} = \dfrac{{2a{t_1} - 0}}{{at_1^2 - a}}$

Now simplify this we have,

$ \Rightarrow \dfrac{{ - t}}{{1 - {t^2}}} = \dfrac{{{t_1}}}{{t_1^2 - 1}}$

$ \Rightarrow - t\left( {t_1^2 - 1} \right) = {t_1}\left( {1 - {t^2}} \right)$

$ \Rightarrow - tt_1^2 + t = {t_1} - {t_1}{t^2}$

$ \Rightarrow {t_1}{t^2} - tt_1^2 = {t_1} - t$

$ \Rightarrow - {t_1}t\left( {{t_1} - t} \right) = {t_1} - t$

$ \Rightarrow - {t_1}t = 1$

$ \Rightarrow - {t_1} = \dfrac{1}{t}$

So coordinate A is $\left( {a{t^2},2at} \right)$, and A’ = $\left( {\dfrac{a}{{{t^2}}},\dfrac{{ - 2a}}{t}} \right)$

Now as we know that the distance between two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as,

$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Let A = $\left( {{x_1},{y_1}} \right)$ = $\left( {a{t^2},2at} \right)$, A’ = $\left( {{x_2},{y_2}} \right)$ = $\left( {\dfrac{a}{{{t^2}}},\dfrac{{ - 2a}}{t}} \right)$

So the length of the focal chord is,

$ \Rightarrow L = \sqrt {{{\left( {\dfrac{a}{{{t^2}}} - a{t^2}} \right)}^2} + {{\left( {\dfrac{{ - 2a}}{t} - 2at} \right)}^2}} $

Now simplify this we have,

$ \Rightarrow L = \sqrt {{a^2}{{\left( {\dfrac{1}{{{t^2}}} - {t^2}} \right)}^2} + 4{a^2}{{\left( {\dfrac{1}{t} + t} \right)}^2}} $

$ \Rightarrow L = \sqrt {{a^2}{{\left( {\dfrac{1}{t} + t} \right)}^2}{{\left( {\dfrac{1}{t} - t} \right)}^2} + 4{a^2}{{\left( {\dfrac{1}{t} + t} \right)}^2}} $

$ \Rightarrow L = \sqrt {{a^2}{{\left( {\dfrac{1}{t} + t} \right)}^2}\left[ {{{\left( {\dfrac{1}{t} - t} \right)}^2} + 4} \right]} $

$ \Rightarrow L = \sqrt {{a^2}{{\left( {\dfrac{1}{t} + t} \right)}^2}\left[ {{{\left( {\dfrac{1}{t} + t} \right)}^2}} \right]} $

$ \Rightarrow L = a{\left( {\dfrac{1}{t} + t} \right)^2}$................ (1)

Now find out the equation of chord AA’.

A = $\left( {{x_1},{y_1}} \right)$ = $\left( {a{t^2},2at} \right)$, A’ = $\left( {{x_2},{y_2}} \right)$ = $\left( {\dfrac{a}{{{t^2}}},\dfrac{{ - 2a}}{t}} \right)$

So the equation of chord is,

$ \Rightarrow y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$

Now substitute the values we have,

$ \Rightarrow y - 2at = \dfrac{{\dfrac{{ - 2a}}{t} - 2at}}{{\dfrac{a}{{{t^2}}} - a{t^2}}}\left( {x - a{t^2}} \right)$

Now simplify it we have,

$ \Rightarrow y - 2at = \dfrac{{ - 2a\left( {\dfrac{1}{t} + t} \right)}}{{a\left( {\dfrac{1}{t} - t} \right)\left( {\dfrac{1}{t} + t} \right)}}\left( {x - a{t^2}} \right)$

\[ \Rightarrow \left( {y - 2at} \right)a\left( {\dfrac{1}{t} - t} \right) = - 2a\left( {x - a{t^2}} \right)\]

\[ \Rightarrow ya\left( {\dfrac{1}{t} - t} \right) - 2{a^2}t\left( {\dfrac{1}{t} - t} \right) = - 2ax + 2{a^2}{t^2}\]

\[ \Rightarrow ya\left( {\dfrac{1}{t} - t} \right) - 2{a^2} + 2{a^2}{t^2} = - 2ax + 2{a^2}{t^2}\]

\[ \Rightarrow ya\left( {\dfrac{1}{t} - t} \right) - 2{a^2} = - 2ax\]

\[ \Rightarrow 2ax + ya\left( {\dfrac{1}{t} - t} \right) - 2{a^2} = 0\]

So this is the equation of focal chord AA’.

Now as we know that the perpendicular drawn from the point $\left( {{x_1},{y_1}} \right)$ on the line ax + by + c = 0 is given as,

$d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

So the distance P from the vertex (0, 0) on the equation of chord is given as

$ \Rightarrow p = \dfrac{{\left| {0 + 0 - 2{a^2}} \right|}}{{\sqrt {{{\left( {2a} \right)}^2} + {a^2}{{\left( {\dfrac{1}{t} - t} \right)}^2}} }}$

$ \Rightarrow p = \dfrac{{2{a^2}}}{{a\sqrt {4 + {{\left( {\dfrac{1}{t} - t} \right)}^2}} }}$

$ \Rightarrow p = \dfrac{{2a}}{{\sqrt {4 + \left( {\dfrac{1}{{{t^2}}} + {t^2} - 2} \right)} }}$

$ \Rightarrow p = \dfrac{{2a}}{{\sqrt {\dfrac{1}{{{t^2}}} + {t^2} + 2} }}$

$ \Rightarrow p = \dfrac{{2a}}{{\sqrt {{{\left( {\dfrac{1}{t} + t} \right)}^2}} }} = \dfrac{{2a}}{{\left( {\dfrac{1}{t} + t} \right)}}$

$ \Rightarrow \left( {\dfrac{1}{t} + t} \right) = \dfrac{{2a}}{p}$

Now substitute this value in equation (1) we have,

$ \Rightarrow L = a{\left( {\dfrac{1}{t} + t} \right)^2} = a{\left( {\dfrac{{2a}}{p}} \right)^2}$

$ \Rightarrow L = \dfrac{{4{a^3}}}{{{p^2}}}$

So this is the required length of the focal chord.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the perpendicular drawn from the point $\left( {{x_1},{y_1}} \right)$ on the line ax + by + c = 0 is given as, $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ and always recall that the distance between two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as, $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $