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Differential Equation of a Curve from Tangent and Normal | JEE

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Question
A tangent to a curve intersects the yy-axis at a point PP. A line perpendicular to this tangent through PP passes through another point (1,0)(1, 0). The differential equation of the curve is:
Aydydx−x(dydx)2=1y\dfrac{dy}{dx} - x\left(\dfrac{dy}{dx}\right)^2 = 1correct
Bxd2ydx2+(dydx)2=1x\dfrac{d^2y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 = 1
Cydxdy+x=1y\dfrac{dx}{dy} + x = 1
DNone of these
Solution
Let P1=(x,y)P_1 = (x, y) with slope m=dydxm = \dfrac{dy}{dx}. The tangent Y−y=m(X−x)Y - y = m(X - x) meets the yy-axis (X=0X = 0) at
P=(0, y−mx).P = \big(0,\ y - mx\big).
 The perpendicular through PP has slope −1m-\dfrac{1}{m}:
Y−(y−mx)=−1m(X−0).Y - (y - mx) = -\frac{1}{m}(X - 0).
 It passes through (1,0)(1, 0):
0−(y−mx)=−1m⇒y−mx=1m⇒my−m2x=1.0 - (y - mx) = -\frac{1}{m} \Rightarrow y - mx = \frac{1}{m} \Rightarrow my - m^2x = 1.
 Replacing m=dydxm = \dfrac{dy}{dx}:
ydydx−x(dydx)2=1.y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 1.
 Correct answer: (1)
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