Question

The tangent to the curve intersects the Y-axis at point P. A line drawn through point P is perpendicular to this tangent and passes through another point \((1,0)\). The differential equation of the curve is...

• A. \(y\frac{dy}{dx}-x(\frac{dy}{dx})^2 = 1\)
• B. \(x\frac{dy}{dx}-y(\frac{dy}{dx})^2 = 1\)
• C. \(y\frac{dy}{dx}+x = 1\)
• D. \(x\frac{dy}{dx}+y = 1\)

Hint:

Find the Y-intercept \(P\) of the tangent, then use the perpendicular through \(P\) and \((1,0)\).

Answer 1

The Correct Option is A

Step 1: Write both slopes
Tangent slope $m$. Perpendicular slope must be $-1/m$.

Step 2: Use points P and (1,0)
Slope from $P(0,y-xm)$ to $(1,0)$ is $-(y-xm)$. Equate to $-1/m$: $m(y-xm)=1$, i.e. $y\,y'-x\,y'^2=1$, option (A).

Final Answer:
Option (A) is the required differential equation. \[ \boxed{y\,y'-x\,(y')^2=1} \]