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Differential Equation of a Curve from Normal and Area | JEE

JEE Maths question with a full step-by-step solution.

Question
A normal at any point (x,y)(x, y) to the curve y=f(x)y = f(x) cuts a triangle of unit area with the axes. The differential equation of the curve is:
Ay2−x2(dydx)2=4dydxy^2 - x^2\left(\dfrac{dy}{dx}\right)^2 = 4\dfrac{dy}{dx}
Bx2−y2(dydx)2=dydxx^2 - y^2\left(\dfrac{dy}{dx}\right)^2 = \dfrac{dy}{dx}
Cx+ydydx=yx + y\dfrac{dy}{dx} = y
Dx2+2xydydx+y2(dydx)2=2dydxx^2 + 2xy\dfrac{dy}{dx} + y^2\left(\dfrac{dy}{dx}\right)^2 = 2\dfrac{dy}{dx}correct
Solution
Step 1: With m=dydxm = \dfrac{dy}{dx}, the normal at (x,y)(x, y) is Y−y=−1m(X−x)Y - y = -\dfrac{1}{m}(X - x). Its intercepts are
X-intercept=x+my,Y-intercept=y+xm.X\text{-intercept} = x + my, \qquad Y\text{-intercept} = y + \frac{x}{m}.
 Step 2: The triangle with the axes has unit area:
12(x+my)(y+xm)=1.\frac{1}{2}\left(x + my\right)\left(y + \frac{x}{m}\right) = 1.
 Step 3: Expand and multiply by mm:
2xy+x2m+my2=2⇒x2+2xy m+y2m2=2m.2xy + \frac{x^2}{m} + my^2 = 2 \Rightarrow x^2 + 2xy\,m + y^2 m^2 = 2m.
 Replacing m=dydxm = \dfrac{dy}{dx}:
x2+2xydydx+y2(dydx)2=2dydx.x^2 + 2xy\frac{dy}{dx} + y^2\left(\frac{dy}{dx}\right)^2 = 2\frac{dy}{dx}.
 Correct answer: (4)
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