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Differential Equation of a Curve Using the Normal | JEE

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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 xx-axis, and passing through (2,1)(2, 1), is:
Ax2+y2−x=0x^2 + y^2 - x = 0
B4x2+2y2−9y=04x^2 + 2y^2 - 9y = 0
C2x2+4y2−9x=02x^2 + 4y^2 - 9x = 0
D4x2+2y2−9x=04x^2 + 2y^2 - 9x = 0correct
Solution
Step 1: The normal at (x,y)(x, y) meets the xx-axis at X=x+ydydxX = x + y\dfrac{dy}{dx}. The condition "square of the ordinate = twice (abscissa ×\times intercept)" gives
y2=2x(x+ydydx)⇒dydx=y2−2x22xy.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=vxy = vx, dydx=v+xdvdx\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}:
v+xdvdx=v2−22v⇒xdvdx=−v2+22v⇒2v dvv2+2=−dxx.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⁡(v2+2)=−ln⁡x+ln⁡c⇒x(v2+2)=c⇒2x+y2x=c⇒y2+2x2=cx.\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)(2, 1): 1+8=2c⇒c=921 + 8 = 2c \Rightarrow c = \dfrac{9}{2}, so 2y2+4x2=9x2y^2 + 4x^2 = 9x, i.e.
4x2+2y2−9x=0.4x^2 + 2y^2 - 9x = 0.
 Correct answer: (4)
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