Question

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

• A. ${\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2=2$
• B. $2y-3x=0$
• C. $y=\frac{6}{x}$
• D. $x^2+y^2=13$

Answer 1

The correct option is C $y=\frac{6}{x}$

Let $B$ be the pt. of contact $y=f\left(x\right)$ be the curve

then $AB=BC$ (given condition)

$C\equiv (2x,0)$ $A\equiv (0,2y)$

Slope m of line AC is

$m=\frac{dy}{dx}=\frac{0.2y}{2x-0}$

$\frac{dy}{dx}=\frac{-y}{x}$

$\frac{dy}{y}=\frac{-dx}{x}$

$lny=-lnx+c$

$yx=c$

as (2,3) satisfies the curve

$\therefore 2\times 3=c$

$\therefore y=\frac{6}{x}$ is the curve

$\therefore$ Option C is correct