Sol. Let y = f(x) and y= g(x) be the required curves. The equation of the tangents to these two curves at points with equal abscissae x(say) are

Y – f(x) = f ′ (x) (X –x) and Y –g(x) = g ′ (x) (X –x)

These two lines intersect at a point on Y axis. Therefore,

Y –f(x) = –xf ′ (x) and Y – g(x) = –xg ′ (x)

⇒ Y = f(x) –xf ′ (x) and Y = g(x) –xg ′ (x)

⇒ f(x) –x f ′ (x) = g(x) – xg ′ (x)

⇒ f(x) –g(x) = x {f ′ (x) – g ′ (x)}

⇒ f(x) – g(x) = x {f(x) –g(x)}

⇒ =

On integrating, we obtain

log {f(x) –g(x)} = log x + log C

⇒ f(x) –g(x) = Cx ... (i)

The equations of the normals at points with equal abscissae x(say) to the two curves are

Y –f(x) = – (X – x) and Y–g(x)

= – (X –x)

These two intersect at a point on X-axis. Therefore,

0 –f(x) = – (X –x) and 0 –g(x)

= – (X –x)

⇒ X = x + f(x) f ′ (x) and X = x + g(x) g ′ (x)

⇒ x + f(x) f ′ (x) = x + g(x) g ′ (x)

⇒ f(x) f ′ (x) = g(x) g ′ (x)

On integrating, we obtain

{f(x)} 2 = {g(x)} 2 + C 1

{f(x)} 2 –{g(x)} 2 = C 1

⇒ {f(x) + g(x)} (Cx) = C 1 [Using (i)]

⇒ f(x) + g(x) = ... (ii)

Solving (i) and (ii), we obtain

f(x) = ... (iii)

and, g(x) = ... (iv)

It is given that (iii) passes through (1, 1) and (iv) passes through (2, 3) therefore,

2 = C + and 6 = –2C

⇒ C = –2 and C 1 = –8

Substituting the values of C 1 and C in (iii) and (iv), we get

f(x) = and g(x) = or, f(x) = –x

and g(x) = x +