# Find the locus of the middle points of the chords for the parabola${y^2} = 4x$ , chord which touches the parabola${y^2} + 4bx = 0$;$\left( {b > 0} \right)$

Last updated date: 23rd Mar 2023

•

Total views: 308.4k

•

Views today: 5.90k

Answer

Verified

308.4k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE