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Differential Equation of a Curve from Tangent and Normal | JEE

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

Question

A tangent to a curve intersects the

y

-axis at a point

P

. A line perpendicular to this tangent through

P

passes through another point

(1, 0)

. The differential equation of the curve is:

y\dfrac{dy}{dx} - x\left(\dfrac{dy}{dx}\right)^2 = 1

correct

x\dfrac{d^2y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 = 1

y\dfrac{dx}{dy} + x = 1

DNone of these

Solution

Let

P_1 = (x, y)

with slope

m = \dfrac{dy}{dx}

. The tangent

Y - y = m(X - x)

meets the

y

-axis (

X = 0

) at

P = \big(0,\ y - mx\big).

The perpendicular through

P

has slope

-\dfrac{1}{m}

Y - (y - mx) = -\frac{1}{m}(X - 0).

It passes through

(1, 0)

0 - (y - mx) = -\frac{1}{m} \Rightarrow y - mx = \frac{1}{m} \Rightarrow my - m^2x = 1.

Replacing

m = \dfrac{dy}{dx}

y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 1.

Correct answer: (1)

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