IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ashish_banga

the tangent at any arbitary pint P(x,y) on the curve y = f(x), meets the x and y axis at the points A & B resp. if PA : PB = 2 : 1,then equation of curve is

krish

Given : y=f(x) is the curve

let the arbitrary point be P(h,k)

Slope of tangent at P = f ' (h) = (dy/dx)_(h,k)

Therefore , equation of tangent at (h,k) will be

y-k = f ' (h) (x-h) ----(1)

The tangent meets the x-axis (y=0) at A

therefore putting y=0 in eq 1,

we get the x-coordinate of A

(ie.) A ( h-(k/f ' (h)) , 0)

In the same way putting x=0 in equation 1

B ( 0, k-(h f ' (h)))

Given that PA : PB = 2 : 1

WE GET ,

square of (PA) : square of (PB) = 4 : 1

By distance formula for PA and PB

k² (1+(f ' (h))² )/(f ' (h))² = 4 h² (1+(f ' (h))² )

which implies

(f ' (h))² = k² / 4h²

Taking root and

REPLACING H BY X AND K BY Y as per the arbitrary point P in the question ,

dy/dx = y/2x

which is solved

to get

ln y = (ln x)/4 + c

if u face any problem , plz reply back

krishna.gopal

Good work Krish, but i there are some mistakes.

First if (dy/dx)²=y²/(4x²)

Then dy/dx=y/2x,-y/2x

Secondly

Solution of dy/dx=y/2x is ln(y)=ln(x)/2+c

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