Q. 255.0( 1 Vote )

# Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with the x - axis is twice the abscissa of the point of contact.

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, the equation of the tangent at P(x,y)

Y – y = (X – x)

Now, putting Y = 0

0 – y = (X – x)

X = x – y So, B(x – y ,0)

Given, the distance between the foot of ordinate of the point of contact and the point of intersection of tangent and x - axis = 2x

BC = 2x y = 2x Integrating both sides we have

logx = 2logy + c……(1)

As it passes through (1,2)

So, the point must satisfy the equation above

log1 = 2log2 + c

0 = 2log2 + c

c = – 2log2

Putting the value of c in equation (1)

logx = 2logy – 2log2

logx = 2(logy – log2)   Rate this question :

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