Question

Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x, y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Answer 1

Let P (x, y) be any point on the curve and AB be the tangent to the given curve at P.

P is the midpoint of AB  .....(Given)

∴ Coordinates of A and B are (2x, 0) and (0, 2y) respectively.

∴ Slope of the tangent AB = ${\displaystyle \frac{2y-0}{0-2x}=-\frac{y}{x}}$

∴ ${\displaystyle \frac{\text{d}y}{\text{d}x}=-\frac{y}{x}}$

⇒ ${\displaystyle \frac{\text{d}y}{y}=-\frac{\text{d}x}{x}}$

Integrating both sides, we get

${\displaystyle \int \frac{\text{d}y}{y}=-\int \frac{\text{d}x}{x}}$

⇒ log y = – log x + log c

⇒ log y + log x = log c

⇒ log yx = log c

∴ yx = c

Since, the curve passes through (1, 1)

∴ 1 × 1 = c

∴ c = 1

Hence, the required equation is xy = 1.