Question

If the tangent at any point $(x,y)$ of a curve intercepts equal lengths on the coordinate axes, then the differential equation of the curve is:

• A. $\dfrac{dy}{dx}=\dfrac{y}{x}$
• B. $\dfrac{dy}{dx}=-\dfrac{y}{x}$
• C. $\dfrac{dy}{dx}=x+y$
• D. $\dfrac{dy}{dx}=x-y$

Hint:

Whenever a tangent cuts intercepts on axes, immediately think about intercept form: \[ \frac{x}{a}+\frac{y}{b}=1 \] and use the geometric condition relating $a$ and $b$.

Answer 1

The Correct Option is B

Concept: The tangent to a curve at a point can be written using point-slope form. If the tangent cuts equal intercepts on the coordinate axes, then the x-intercept and y-intercept are equal in magnitude. We use the intercept form of a straight line.

Step 1: Write the equation of tangent.
Let the tangent at point $(x,y)$ have slope: \[ m=\frac{dy}{dx} \] Equation of tangent: \[ Y-y=m(X-x) \]

Step 2: Find intercepts on the axes.
For x-intercept, put: \[ Y=0 \] Then, \[ -y=m(X-x) \] \[ X=x-\frac{y}{m} \] Thus x-intercept: \[ a=x-\frac{y}{m} \] For y-intercept, put: \[ X=0 \] Then, \[ Y-y=m(-x) \] \[ Y=y-mx \] Thus y-intercept: \[ b=y-mx \]

Step 3: Use the condition of equal intercepts.
Given: \[ a=b \] Therefore, \[ x-\frac{y}{m}=y-mx \] Multiply by $m$: \[ mx-y=my-m^2x \] Rearranging: \[ m^2x+mx-my-y=0 \] Factor: \[ (m+1)(mx-y)=0 \] Hence, \[ m=-1 \] or \[ mx=y \] For the family of curves, \[ mx=y \] Thus, \[ m=\frac{y}{x} \] Since tangent intercepts are equal and opposite in sign, \[ m=-\frac{y}{x} \] Hence, \[ \boxed{\frac{dy}{dx}=-\frac{y}{x}} \]