Q. 195.0( 1 Vote )

Find the equation of the curve such that the portion of the x - axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer :

Let P(x,y) be the point of contact of tangent and curve y = f(x).

It cuts the axes at A and B so, equation of tangent at P(x,y)


Y – y = (X – x)


Putting X = 0


Y – y = (0 – x)


Y = y – x


So, A(0, y – x)


Now, putting Y = 0


0 – y = (X – x)


X = x – y


So, B(x – y,0)


Given, intercept on x - axis = 4× ordinate


x – y = 2x


– y = x



– logx = logy + c ……(1)


As it passes through (1,2)


So, the point must satisfy the equation above


– log1 = log2 + c


0 = log2 + c


c = – log2


Putting the value of c in equation (1)


– logx = logy – log2


log2 = logx + logy


log2 = logxy


xy = 2


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