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: |
$y=\frac{6}{x}$ $x^2+y^2=13$ $\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\right)^2=2$ $2 y-3 x=0$ |
$y=\frac{6}{x}$ |
Let P(x, y) be any point on the curve. The equation of tangent at P is given by $Y-y=\frac{d y}{d x}(X-x)$ This cuts coordinate axes at $A\left(x-y \frac{d x}{d y}, 0\right)$ and $B\left(0, y-x \frac{d y}{d x}\right)$. It is given that $P(x, y)$ is the mid-point of $A B$. $2 x=x-y \frac{d x}{d y}$ and $2 y=y-x \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x}=-\frac{y}{x}$ $\Rightarrow x d y+y d x=0 \Rightarrow d(x y)=0 \Rightarrow x y=C$ The curve passes through $(2,3)$. ∴ $2 \times 3=C \Rightarrow C=6$ The equation of the curve is $x y=6$ or, $y=\frac{6}{x}$. |