UPSC 2014 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Variables separable · ODEs · asked 3× in 13 yrs · Read the full method →

Question

Find the curve for which the part of the tangent cut-off by the axes is bisected at the point of tangency.

Geometric tangent-intercept formula; midpoint condition gives ODE; separation of variables.

Strategy. Express the geometric condition (point of tangency is midpoint of tangent’s axis intercepts) as an ODE; solve.

At point $(x_0,y_0)$ on the curve $y=f(x)$ with slope $f'(x_0)$ , the tangent line is

y-y_0=f'(x_0)(x-x_0).

$x$ -intercept (set $y=0$ ): $-y_0=f'(x_0)(x-x_0)$ , so $x=x_0-y_0/f'(x_0)$ .
$y$ -intercept (set $x=0$ ): $y=y_0-x_0\,f'(x_0)$ .

The point of tangency $(x_0,y_0)$ is the midpoint of the segment between the two intercepts:

x_0=\dfrac{(x_0-y_0/f'(x_0))+0}{2},\qquad y_0=\dfrac{0+(y_0-x_0\,f'(x_0))}{2}.

From the $x$ -equation: $2x_0=x_0-y_0/f'(x_0)\Rightarrow x_0=-y_0/f'(x_0)\Rightarrow f'(x_0)=-y_0/x_0$ .

(The $y$ -equation gives the same condition.)

Drop the subscripts:

\dfrac{dy}{dx}=-\dfrac{y}{x}.

Separable: $\dfrac{dy}{y}=-\dfrac{dx}{x}$ .

Integrate: $\ln|y|=-\ln|x|+C_1$ , i.e., $\ln|xy|=C_1$ , $xy=e^{C_1}=$ const.

\boxed{\;xy=k\quad(k\text{ a non-zero constant — rectangular hyperbola}).\;}