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Differential Equation of a Curve Using the Normal | JEE

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

Question

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on the

x

-axis, and passing through

(2, 1)

, is:

x^2 + y^2 - x = 0

4x^2 + 2y^2 - 9y = 0

2x^2 + 4y^2 - 9x = 0

4x^2 + 2y^2 - 9x = 0

correct

Solution

Step 1: The normal at

(x, y)

meets the

x

-axis at

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

. The condition "square of the ordinate = twice (abscissa

\times

intercept)" gives

y^2 = 2x\left(x + y\frac{dy}{dx}\right) \Rightarrow \frac{dy}{dx} = \frac{y^2 - 2x^2}{2xy}.

Step 2: This is homogeneous; put

y = vx

\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}

v + x\frac{dv}{dx} = \frac{v^2 - 2}{2v} \Rightarrow x\frac{dv}{dx} = -\frac{v^2 + 2}{2v} \Rightarrow \frac{2v\,dv}{v^2 + 2} = -\frac{dx}{x}.

Step 3: Integrate:

\ln(v^2 + 2) = -\ln x + \ln c \Rightarrow x(v^2 + 2) = c \Rightarrow 2x + \frac{y^2}{x} = c \Rightarrow y^2 + 2x^2 = cx.

Step 4: Through

(2, 1)

1 + 8 = 2c \Rightarrow c = \dfrac{9}{2}

, so

2y^2 + 4x^2 = 9x

, i.e.

4x^2 + 2y^2 - 9x = 0.

Correct answer: (4)

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