Question

The curve that passes through the point (2,3) and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

Answer 1

Let the point (x, y) is on the curve , and slope of tangent of curve is dy/dx [ slope of tangent of curve f(x) = y is always df(x)/dx /dy/dx ]
so, equation of tangent is given by
(X - x) = dy/dx (Y - y)
So, X - intercept of tangent ⇒ (X - x) = dy/dx (0 - y)
⇒X - x = -ydy/dx
⇒X = x - ydy/dx [ it is X - intercept ]

similarly Y - intercept of tangent ⇒(0 - x) = dy/dx(Y - y)
⇒-x = Y.dy/dx - ydy/dx
⇒ ydy/dx - x = Ydy/dx
⇒y - x.dx/dy = Y [ it is Y- intercept ]

Now, According to question,
X = x - ydy/dx = 2x and Y = y - xdx/dy = 2y

x - ydy/dx = 2x
⇒ - ydy/dx = 2x - x = x
⇒ dy/dx = -y/x
⇒ dy/y = -dx/x
⇒∫dy/y + ∫dx/x = 0
Integrating
⇒lny + lnx = lnC [ C is constant ]
⇒ lnyx = lnC
⇒ yx = C ⇒ y = C/x

Now, (2,3) lies on the y = C/x curve so, 3 = C/2 ⇒C = 6
Hence, curve is y = 6/x