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# Find the locus of the middle points of the chords for parabola ${y^2} = 4x$ chords which are of length $\;2b$.

Last updated date: 17th Mar 2023
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Hint: Using the slope-point form, find the equation of the tangent and the chord of contact. Using the formula $L = \dfrac{4}{{{m^2}}}\sqrt {a\left( {1 + {m^2}} \right)\left( {a - mc} \right)}$ to find the length of the chord, and equating it to $\;2b$, find the locus.

The given equation of the parabola is
${y^2} = 4x$ …(1)
Let P and Q be points on the parabola. PQ is a chord for the parabola ${y^2} = 4x$
Length of the chord PQ=$\;2b$
Let $M\left( {h,k} \right)$be the midpoint of the chord PQ of the parabola.
At$M\left( {h,k} \right)$, equation (1) becomes
${k^2} = 4ah$
${k^2} - 4ah = 0$ …(2)
Slope Point form to find the equation of a line passing through a point $\left( {x_1,y_1} \right)$ with a slope $m$ is written as $\left( {y - y_1} \right) = m\left( {x - x_1} \right)$
We get the slope by differentiating ${y^2} = 4ax$ with respect to $x$.
So, let us differentiate equation (1) with respect to $x$ to find its slope.
${y^2} = 4ax \\ 2y\dfrac{{dy}}{{dx}} = 4a \\ \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} \\$
$m = \dfrac{{2a}}{{y_1}}$
Equation of tangent to the parabola ${y^2} = 4ax$ at any point $A\left( {x_1,y_1} \right)$ is given by the slope-point form as
$y - y_1 = m\left( {x - x_1} \right) \\ y - y_1 = \dfrac{{2a}}{{y_1}}\left( {x - x_1} \right) \\$
$yy_1 - y{1^2} = 2ax - 2ax_1$ …(3)
${y^2} = 4ax$at$A\left( {x_1,y_1} \right)$ is
$y{1^2} = 4ax_1$ …(4)
Substitute (4) in (3),
$yy_1 - y{1^2} = 2ax - 2ax_1 \\ yy_1 - 4ax_1 = 2ax - 2ax_1 \\ yy_1 = 2a\left( {x + x_1} \right) \\$
Since, T=0 for the chord of contact, we get the equation for chord of contact as $yy_1 - 2a\left( {x + x_1} \right) = 0$
Hence, the equation for chord of contact is
$yy_1 - 2a\left( {x + x_1} \right) = 0$ …(5)
At $M\left( {h,k} \right)$, equation (5) becomes
$ky - 2ax - 2ah = 0$ …(6)
Equating equations (2) and (6) we get
${k^2} - 4ah = ky - 2ax - 2ah$
${k^2} - 2ah = ky - 2ax$ …(7)
Rearranging equation (7) in the form $y = mx + c$
${k^2} - 2ah = ky - 2ax \\ ky = {k^2} - 2ah + 2ax \\ y = \dfrac{{2a}}{k}x + \dfrac{{{k^2} - 2ah}}{k} \\$
Comparing with $y = mx + c$, we get
$m = \dfrac{{2a}}{k},c = \dfrac{{{k^2} - 2ah}}{k}$ …(8)
Length of the chord for the parabola ${y^2} = 4x$ is given by
$L = \dfrac{4}{{{m^2}}}\sqrt {a\left( {1 + {m^2}} \right)\left( {a - mc} \right)}$
Length of the chord is given as$\;2b$.
Substituting (8) in this we get,
$2b = \dfrac{4}{{{{\left( {\dfrac{{2a}}{k}} \right)}^2}}}\sqrt {a\left( {1 + \dfrac{{4{a^2}}}{{{k^2}}}} \right)\left[ {a - \left( {\dfrac{{2a}}{k}} \right)\left( {\dfrac{{{k^2} - 2ah}}{k}} \right)} \right]} \\ 2b = \dfrac{{{k^2}}}{{{a^2}}}\sqrt {a\left( {\dfrac{{{k^2} + 4{a^2}}}{{{k^2}}}} \right)\left( {\dfrac{{a{k^2} - 2a{k^2} + 4{a^2}h}}{{{k^2}}}} \right)} \\$
Squaring on both sides,
$4{a^4}{b^2} = \dfrac{{{k^4}}}{{{k^4}}}\left[ {a\left( {{k^2} + 4{a^2}} \right)\left( {a{k^2} - 2a{k^2} + 4{a^2}h} \right)} \right] \\ 4{a^3}{b^2} = \left( {{k^2} + 4{a^2}} \right)a\left( {2ah - {k^2}} \right) \\ 4{a^2}{b^2} = \left( {{k^2} + 4{a^2}} \right)\left( {2ah - {k^2}} \right) \\$
When we replace the point $\left( {h,k} \right)$by$\left( {x,y} \right)$, we will get the locus of the middle points of the chord of length $2b$ to the parabola ${y^2} = 4x$
$4{a^2}{b^2} = \left( {{y^2} + 4{a^2}} \right)\left( {2ax - {y^2}} \right)$ is the required locus.
Note: Memorizing the formula for length of the chord and the equation of the chord will make this very simple instead of deriving it each time. Finding the slope and intercept from the equation of the chord and substituting it in the length of the chord formula gives the locus of the middle points of the chord of length $2b$ to the parabola ${y^2} = 4x$.