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: Understanding the Concept
Let the curve be \(y=y(x)\) with slope \(m=\dfrac{dy}{dx}\) at a point \((x,y)\).

Step 2: Key Formula or Approach
Tangent at \((x,y)\) meets the Y-axis where its equation \(Y-y=m(X-x)\) gives \(X=0\): \(P=(0,\,y-xm)\).

Step 3: Detailed Explanation
The line through \(P\) and \((1,0)\) has slope \(\dfrac{0-(y-xm)}{1-0}=xm-y\).
It is perpendicular to the tangent, so \(m(xm-y)=-1\).
\[ xm^2-ym=-1 \Rightarrow ym-xm^2=1 \]
\[ y\frac{dy}{dx}-x\left(\frac{dy}{dx}\right)^2=1 \]

Final Answer:
The differential equation is \(y\frac{dy}{dx}-x\left(\frac{dy}{dx}\right)^2=1\), option (A). \[ \boxed{y\dfrac{dy}{dx}-x\left(\dfrac{dy}{dx}\right)^2=1\ \text{(A)}} \]