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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)

to the curve

y = f(x)

cuts a triangle of unit area with the axes. The differential equation of the curve is:

y^2 - x^2\left(\dfrac{dy}{dx}\right)^2 = 4\dfrac{dy}{dx}

x^2 - y^2\left(\dfrac{dy}{dx}\right)^2 = \dfrac{dy}{dx}

x + y\dfrac{dy}{dx} = y

x^2 + 2xy\dfrac{dy}{dx} + y^2\left(\dfrac{dy}{dx}\right)^2 = 2\dfrac{dy}{dx}

correct

Solution

Step 1: With

m = \dfrac{dy}{dx}

, the normal at

(x, y)

Y - y = -\dfrac{1}{m}(X - x)

. Its intercepts are

X\text{-intercept} = x + my, \qquad Y\text{-intercept} = y + \frac{x}{m}.

Step 2: The triangle with the axes has unit area:

\frac{1}{2}\left(x + my\right)\left(y + \frac{x}{m}\right) = 1.

Step 3: Expand and multiply by

m

2xy + \frac{x^2}{m} + my^2 = 2 \Rightarrow x^2 + 2xy\,m + y^2 m^2 = 2m.

Replacing

m = \dfrac{dy}{dx}

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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