IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Mr.IITIAN007

Dear Experts !
What is the Cartesian equation of the curves for which length of the tangent is equal to the radius vector ?

nitin62225

ans: xy=c or x/y =c

soln:

radius vector's modulous= (x²+y²)^1/2

and length of tangent = IyI(1+(dx/dy)²)^1/2

so u get a diff. equn, dx/dy= (+,-)x/y

nd on solving u get xy=c or x/y=c as the required answer.

naveen_nvn

if tan a = dy/dx b the slope of tangent...
y sin a is the length of tangent...and the radius vector is
(x^2 + y^2)^0.5...equating these two and expressing sin a in term sof tan a i.e, dy/dx u get the required soln...but the differential eqn obtained will not b of 1 st order so its no thr in jee syllabus i guess...
cheers

naveen_nvn

dear nitin...u have taken length of tangent as y sec a....but its actually y sin a...and hence u wont get a simple differential eqn like u have got..

Mr.IITIAN007

ok guys ...I am waiting for the experts to answer .If your respective answers are correct I will definitely Rate.

iitkgp_bipin

Equation of any tangent at any point P(x,y) is Y-y = (dy/dx)(X-x)
which intersects x-axis at T(x - (dx/dy)y , 0).

Length of tangent, PT = {(dx/dy)²y² + y²}^1/2
Length of radius = (x²+y²)^1/2

Equating these two and squaring :
{(dx/dy)²y² + y²} = (x²+y²)
(dx/dy)²y² = x²
(dy/y) =

(dx/x)

Taking +ve sign and integrating :
y = c₁x (c₁ is any constant)

Taking -ve sign :
y = c₂/x (c₂ is other constant)

Mr.IITIAN007

Thanx bipin.

Ask & Discuss Questions with Community & Experts