Question

find the orthogonal trajectory of the family of circlesx2+y2+2gx+C =0 where 'g' is a parameter​

Answer 1

$\large\underline{\sf{Solution-}}$

The given curve is

$\bf :\longmapsto\: {x}^{2} + {y}^{2} + 2gx + c = 0 - - - (1)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:2x + 2y\dfrac{dy}{dx} + 2g = 0$

$\rm :\longmapsto\:x + y\dfrac{dy}{dx} + g = 0$

$\rm :\longmapsto\: - x - y\dfrac{dy}{dx} = g$

$\bf :\longmapsto\: g \: = \: - (x + y\dfrac{dy}{dx}) - - - (2)$

On substituting the value of g in equation (1), we get

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2x\bigg(x + \dfrac{dy}{dx}\bigg) + c = 0$

Now,

$\bf :\longmapsto\:Replace \: \dfrac{dy}{dx} = - \: \dfrac{dx}{dy}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 2x\bigg(x - \dfrac{dx}{dy}\bigg) + c = 0$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - {2x}^{2} + 2xy\dfrac{dx}{dy} + c = 0$

$\rm :\longmapsto\: - {x}^{2} + {y}^{2}+ 2xy\dfrac{dx}{dy} + c = 0$

$\rm :\longmapsto\: - {x}^{2}+ 2xy\dfrac{dx}{dy} = - c - {y}^{2}$

$\rm :\longmapsto\: 2xy\dfrac{dx}{dy} - {x}^{2} = - c - {y}^{2}$

Divide both sides by y, we get

$\rm :\longmapsto\:2x\dfrac{dx}{dy} - \dfrac{ {x}^{2} }{y} = - y - \dfrac{c}{y} - - - (2)$

$\red{\rm :\longmapsto\:Put \: {x}^{2} = t} \\ \red{\rm :\longmapsto\:2x\dfrac{dx}{dy} = \dfrac{dt}{dy}}$

So, equation (2) can be rewritten as

$\rm :\longmapsto\:\dfrac{dt}{dy} - \dfrac{t}{y} = - y - \dfrac{c}{y}$

☆ which is the form of Linear Differential equation

☆ On comparing with,

$\bf :\longmapsto\:\dfrac{dt}{dy} + pt = q$

☆ On comparing we get,

$\red{\rm :\longmapsto\:p = - \dfrac{1}{y}} \\ \red{\rm :\longmapsto\:q = - y - \dfrac{c}{y}}$

Therefore,

☆ Integrating Factor is evaluated as

$\rm :\longmapsto\:I.F. = {e}^{ \displaystyle \int \sf \: pdy}$

$\rm :\longmapsto\:I.F. = {e}^{ \displaystyle \int \sf \: - \dfrac{1}{y} dy}$

$\rm :\longmapsto\:I.F. = {e}^{ - log(y) }$

$\rm :\longmapsto\:I.F. = {e}^{log {y}^{ - 1} }$

$\bf\implies \:I.F. = {y}^{ - 1} = \dfrac{1}{y}$

Hence,

☆ Solution of equation is given by

$\rm :\longmapsto\:t \times I.F. = \displaystyle \int \sf\: (q \times I.F.)dy$

$\rm :\longmapsto\:t \times \dfrac{1}{y} = \displaystyle \int \sf\: \bigg( - y - \dfrac{c}{y} \bigg) \times \dfrac{1}{y} \: \: dy$

$\rm :\longmapsto\:\dfrac{t}{y} = \displaystyle \int \sf\: \bigg( - 1- \dfrac{c}{ {y}^{2} } \bigg)dy$

$\rm :\longmapsto\:\dfrac{t}{y} = - y + \dfrac{c}{y} + d$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{y} = - y + \dfrac{c}{y} + d$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{2} = t \bigg \}}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{y} = \dfrac{ - {y}^{2} + c + dy }{y}$

$\rm :\longmapsto\: {x}^{2} = - {y}^{2} + c + dy$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - c - dy = 0 \: where \: d \: is \: arbitrary \: constant.$

Answer 2

Answer:

Given curve is

• x^+y^+2gx+c=0__(1)eq

differentiating w.r.t.x.we have

• 2x+2ydy/dx+2g=0
• 2(x+ydy/dx+g)=0
• x+ydy/dx+g=0/2
• x+ydy/dx+g=0
• g=-x-ydy/dx
• g=-(x+ydy/dx)__(2)eq

subtutiting (1)eq in (2)eq

• x^+y^+2(-(x+ydy/dx))x+c=0
• x^+y^-2x^_2xydy/dx+c=0
• y^-x^-2xydy/dx+c=0

This is differential equation of eq(1),

we replace,(dy/dx=-dx/dy)

• y^-x^-2xy(-dx/dy)+c=0
• y^-x^+2xydx/dy+c=0
• 2xydx/dy-x^=-c-y^

Dividing "y" on both sides

• 2xdx/dy-x^/y=-(c+y^/y)__(3)eq

let x^=z

D.w.r.t."x".o.b.s

2xdx/dy=dz/dy

• (3)__dz/dy-1/y.x^=-(c+y^/y)

This is in the from of

( dz/dy+pz=q )

Here,p=-1/y;q=-(c+y^/y)

l.F=e~pdy

=e~-1/ydy

=e-logy

=elogy

l.F=1/y

General solution is

• (I.F)z=~q(I.F)dy+c1
• 1/y.z=~-(c+y^/y)(1/y)dy+c1
• x^/y=-~(c+y^/y^)dy+c1
• x^/y=-~(c/y^+1)dy+c1
• x^/y=-~(1/y^.c+1)dy+c1
• x^/y=-~(y*2.c+1)dy+c1
• x^/y=-y*-2+1/-2+1.c-y+c1
• x^/y=-y*-1/-1.c-y+c1
• x^/y=y*-1.c-y+c1
• x^/y=1/y.c-y+c1
• x^/y=c/y-y+c1
• x^/y=c-y^+c1y/y
• x^=c-y^+c1y
• x^+y^-c-c1y=0

The orthogonal trajectory is

• {x^+y^-c-c1y=0}