Find a pair of curves such that (a) the tangents drawn at points with equal abscissas intersect on the y-axis. (b) the normal drawn at points with equal abscissas intersect on the x-axis. (c) one curve passes through (1,1) and other passes through (2, 3).

Explanation

Let the curve be `y=f_(1)(x)` and `y=f_(2)(x)`. Equations of tangents with equal abscissa x are
`Y=f_(1)(x) = f^(')(x)(X-x)` and `Y-f_(2)(x)=f_(2)^(')(X-x)`
These tangent increase at y-axis.
So, their y-intercept are same.
`therefore -xf_(1)^(')(x) + f_(1)(x) = -xf_(2)^(')(x) + f_(2)^(')(x)`
or `f_(1)(x) - f_(2)(x) = x(f_(1)^(')(x) - f_(2)^(')(x))`
or `int(f_(1)^(')(x)-f_(2)^(')(x))/(f_(1)(x)-f_(2)(x)) = int(dx)/(x)`
or `"ln"|f_(1)(x)-f_(2)(x)|="ln"|x|+"ln"C_(1)`
Now, equations of normal with equal abscissa x are
`(Y-f_(1)(x)) = -1/(f_(1)(x))(X-x)`
and `(Y-f_(2)(x)) = -1/(f_(2)^(')(x)) (X-x)`
As these normals intersect on the x-axis,
`x+f_(1)(x)f_(1)^(')(x) = x+f_(2)(x). f_(2)^(')(x)`
or `f_(1)(x)f_(1)^(')(x)-f_(x)f_(2)^(')(x)=0`
Integrating, we get `f_(1)^(2)(x) - f_(1)^(2) = C_(2)`
or `f_(1)(x) + f_(2)(x) = C_(2)/(f_(1)(x) - f_(2)(x))`
`=+-(C_(2)/(C_(1)(x))) = +-(lambda_(2))/(x)`..............(2)
From equations (1) and (2), we get
`2f_(1)(x) = +- (lambda_(2)/x+C_(1)x)`,
`2f_(2)(x)=+-(lambda_(2)/x-C_(1)x)`
We have, `f_(1)(1)=1` and `f_(2)=3`
`therefore f_(1)(x) =2/x-x, f_(2)(x)=2/x+x`