Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

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:

Options:

$y=\frac{6}{x}$

$x^2+y^2=13$

$\left(\frac{x}{2}\right)^2+\left(\frac{y}{3}\right)^2=2$

$2 y-3 x=0$

Correct Answer:

$y=\frac{6}{x}$

Explanation:

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, $y=\frac{6}{x}$.