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Differential Equation of a Curve from a Tangent | JEE

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

Question

A tangent drawn at any point

P

on a curve meets the

x

-axis at

Q

such that the circumcentre of

\triangle POQ

has abscissa half that of its ordinate. The differential equation of such a curve is:

\dfrac{dy}{dx} = \dfrac{x + 2y}{2x - y}

correct

\dfrac{dy}{dx} = \dfrac{2y - x}{2x + y}

\dfrac{dy}{dx} = \dfrac{2y - x}{2x + y}

DNone of these

Solution

Let

P = (x, y)

O = (0,0)

, and the tangent slope

m = \dfrac{dy}{dx}

. The tangent meets the

x

-axis at

Q = \left(x - \dfrac{y}{m},\ 0\right)

. Step 1: Let the circumcentre be

C = \left(\dfrac{t}{2},\ t\right)

(abscissa half the ordinate). Since

O

and

Q

lie on the

x

-axis, the perpendicular bisector of

OQ

is vertical, so the abscissa of

C

\dfrac{1}{2}\left(x - \dfrac{y}{m}\right) = \dfrac{t}{2}

, giving

t = x - \frac{y}{m}.

Step 2: Use

CO = CP

\frac{t^2}{4} + t^2 = \left(\frac{t}{2} - x\right)^2 + (t - y)^2 \Rightarrow x^2 + y^2 = t(x + 2y).

Step 3: Substitute

t = x - \dfrac{y}{m}

and simplify:

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

\Rightarrow \frac{x + 2y}{m} = 2x - y \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{2x - y}.

Correct answer: (1)

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