Find the locus of the middle points of the chords for the parabola${y^2} = 4x$ , chord which touches the parabola${y^2} + 4bx = 0$;$\left( {b > 0} \right)$
Answer
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Hint: Use the slope-point form$\left( {y - y_1} \right) = m\left( {x - x_1} \right)$to find the equation of the tangent and find the chord of contact. Then use the condition that the discriminant is zero at the point of tangency to find the required locus.
The given equation of the parabola is
${y^2} = 4x$ …(1)
We find that$a = 1$
Let P and Q be points on the parabola. PQ is a chord for the parabola ${y^2} = 4x$
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) and putting $a = 1$ we get the equation of the chord PQ.
${k^2} - 4ah = ky - 2ax - 2ah$
${k^2} - 2h = ky - 2x$
$y = k - \dfrac{{2h}}{k} + \dfrac{{2x}}{k}$ …(7)
The chord touches the parabola
${y^2} + 4bx = 0$ …(8)
Put equation (7) in (8)
$
{\left[ {k - \dfrac{{2h}}{k} + \dfrac{{2x}}{k}} \right]^2} + 4bx = 0 \\
{\left[ {\dfrac{{2x}}{k} + \left( {\dfrac{{{k^2} - 2h}}{k}} \right)} \right]^2} + 4bx = 0 \\
\dfrac{{4{x^2}}}{{{k^2}}} + \dfrac{{{{\left( {{k^2} - 2h} \right)}^2}}}{{{k^2}}} + 2\left( {\dfrac{{2x}}{k}} \right)\left( {\dfrac{{{k^2} - 2h}}{k}} \right) + 4bx = 0 \\
\dfrac{1}{{{k^2}}}\left[ {4{x^2} + {k^2} + 4{h^2} - 4{k^2}h + 4x\left( {{k^2} - 2h} \right)} \right] + 4bx = 0 \\
4{x^2} + {k^2} + 4{h^2} - 4{k^2}h + 4x{k^2} - 8xh = - 4bx{k^2} \\
4{x^2} + 4x\left( {{k^2} - 2h} \right) + 4h\left( {h - {k^2}} \right) + {k^2} = - 4bx{k^2} \\
$
$4{x^2} + 4x\left( {{k^2} - 2h + b{k^2}} \right) + \left( {4{h^2} - 4h{k^2} + {k^2}} \right) = 0$ …(9)
Equation (9) is of the form,$A{x^2} + Bx + C = 0$.
$A = 4,B = 4\left( {{k^2} - 2h + b{k^2}} \right),C = 4{h^2} - 4h{k^2} + {k^2}$
The condition for tangency is that the discriminant ${B^2} - 4AC = 0$
Since, the locus of the middle points of the chords for the parabola ${y^2} = 4x$which touches the parabola ${y^2} + 4bx = 0$ is required, we use this condition for tangency.
$
{\left[ {4\left( {{k^2} - 2h + b{k^2}} \right)} \right]^2} = 16\left( {4{h^2} - 4h{k^2} + {k^2}} \right) \\
16\left( {{k^4} + 4{h^2} + {b^2}{k^4} - 4{k^2}h - 4hb{k^2} + 2b{k^4}} \right) = 16\left( {4{h^2} - 4h{k^2} + {k^2}} \right) \\
{k^4}\left( {1 + {b^2} + 2b} \right) - 4hb{k^2} - {k^2} = 0 \\
{k^4}{\left( {1 + b} \right)^2} - {k^2}\left( {4hb + 1} \right) = 0 \\
{k^2}\left[ {{k^2}{{\left( {1 + b} \right)}^2} - \left( {4hb + 1} \right)} \right] = 0 \\
$
${k^2}{\left( {1 + b} \right)^2} - \left( {4hb + 1} \right) = 0$ …(10)
Replacing points $\left( {h,k} \right)$ by $\left( {x,y} \right)$in the equation (10), we get
${y^2}{\left( {1 + b} \right)^2} - \left( {4bx + 1} \right) = 0$is the required locus.
Note: The equation of the chord is found for the first parabola at the midpoint $\left( {h,k} \right)$and since this touches the other parabola, substitute one value in the other to get an equation. From that equation, use the condition for tangency ($D = 0$) to find the required locus, as it just touches the other parabola.
The given equation of the parabola is
${y^2} = 4x$ …(1)
We find that$a = 1$
Let P and Q be points on the parabola. PQ is a chord for the parabola ${y^2} = 4x$
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) and putting $a = 1$ we get the equation of the chord PQ.
${k^2} - 4ah = ky - 2ax - 2ah$
${k^2} - 2h = ky - 2x$
$y = k - \dfrac{{2h}}{k} + \dfrac{{2x}}{k}$ …(7)
The chord touches the parabola
${y^2} + 4bx = 0$ …(8)
Put equation (7) in (8)
$
{\left[ {k - \dfrac{{2h}}{k} + \dfrac{{2x}}{k}} \right]^2} + 4bx = 0 \\
{\left[ {\dfrac{{2x}}{k} + \left( {\dfrac{{{k^2} - 2h}}{k}} \right)} \right]^2} + 4bx = 0 \\
\dfrac{{4{x^2}}}{{{k^2}}} + \dfrac{{{{\left( {{k^2} - 2h} \right)}^2}}}{{{k^2}}} + 2\left( {\dfrac{{2x}}{k}} \right)\left( {\dfrac{{{k^2} - 2h}}{k}} \right) + 4bx = 0 \\
\dfrac{1}{{{k^2}}}\left[ {4{x^2} + {k^2} + 4{h^2} - 4{k^2}h + 4x\left( {{k^2} - 2h} \right)} \right] + 4bx = 0 \\
4{x^2} + {k^2} + 4{h^2} - 4{k^2}h + 4x{k^2} - 8xh = - 4bx{k^2} \\
4{x^2} + 4x\left( {{k^2} - 2h} \right) + 4h\left( {h - {k^2}} \right) + {k^2} = - 4bx{k^2} \\
$
$4{x^2} + 4x\left( {{k^2} - 2h + b{k^2}} \right) + \left( {4{h^2} - 4h{k^2} + {k^2}} \right) = 0$ …(9)
Equation (9) is of the form,$A{x^2} + Bx + C = 0$.
$A = 4,B = 4\left( {{k^2} - 2h + b{k^2}} \right),C = 4{h^2} - 4h{k^2} + {k^2}$
The condition for tangency is that the discriminant ${B^2} - 4AC = 0$
Since, the locus of the middle points of the chords for the parabola ${y^2} = 4x$which touches the parabola ${y^2} + 4bx = 0$ is required, we use this condition for tangency.
$
{\left[ {4\left( {{k^2} - 2h + b{k^2}} \right)} \right]^2} = 16\left( {4{h^2} - 4h{k^2} + {k^2}} \right) \\
16\left( {{k^4} + 4{h^2} + {b^2}{k^4} - 4{k^2}h - 4hb{k^2} + 2b{k^4}} \right) = 16\left( {4{h^2} - 4h{k^2} + {k^2}} \right) \\
{k^4}\left( {1 + {b^2} + 2b} \right) - 4hb{k^2} - {k^2} = 0 \\
{k^4}{\left( {1 + b} \right)^2} - {k^2}\left( {4hb + 1} \right) = 0 \\
{k^2}\left[ {{k^2}{{\left( {1 + b} \right)}^2} - \left( {4hb + 1} \right)} \right] = 0 \\
$
${k^2}{\left( {1 + b} \right)^2} - \left( {4hb + 1} \right) = 0$ …(10)
Replacing points $\left( {h,k} \right)$ by $\left( {x,y} \right)$in the equation (10), we get
${y^2}{\left( {1 + b} \right)^2} - \left( {4bx + 1} \right) = 0$is the required locus.
Note: The equation of the chord is found for the first parabola at the midpoint $\left( {h,k} \right)$and since this touches the other parabola, substitute one value in the other to get an equation. From that equation, use the condition for tangency ($D = 0$) to find the required locus, as it just touches the other parabola.
Last updated date: 18th Sep 2023
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