JEE Challenger
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MathHardAdv 2019 • Paper 1

Differential Equation of Curve with Constant Tangent Segment Length

Let Γ\Gamma be a curve represented by y=y(x)y = y(x) located in the first quadrant, passing through (1,0)(1, 0). Suppose the tangent to Γ\Gamma at any arbitrary point PP intersects the y-axis at YPY_P. If the length of the line segment PYPPY_P is equal to 11 for every point PP on Γ\Gamma, which of the following statements is/are correct?

Options

A

y=log⁡e(1+1−x2x)−1−x2y = \log_e\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) - \sqrt{1 - x^2}

Correct
B

xy′+1−x2=0xy' + \sqrt{1 - x^2} = 0

Correct
C

y=−log⁡e(1+1−x2x)+1−x2y = -\log_e\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) + \sqrt{1 - x^2}

D

xy′−1−x2=0xy' - \sqrt{1 - x^2} = 0

Topics & Concepts

Differential EquationsDifferential EquationsTangents and Normals

Step-by-Step Solution

To find the equation of the curve Γ\Gamma given by y=y(x)y = y(x), let us consider an arbitrary point P(x,y)P(x, y) on the curve in the first quadrant, where x∈(0,1]x \in (0, 1] and y≥0y \ge 0.

The equation of the tangent line to the curve at the point P(x,y)P(x, y) is given by: Y−y=y′(X−x)Y - y = y'(X - x)

To find the yy-intercept YPY_P, we set X=0X = 0: YP=(0,y−xy′)Y_P = \left(0, y - x y'\right)

The length of the line segment PYPP Y_P is calculated using the distance formula between P(x,y)P(x, y) and YP(0,y−xy′)Y_P(0, y - x y'): Length(PYP)=(x−0)2+(y−(y−xy′))2=x2+x2(y′)2=x1+(y′)2\text{Length}(P Y_P) = \sqrt{(x - 0)^2 + \big(y - (y - x y')\big)^2} = \sqrt{x^2 + x^2 (y')^2} = x \sqrt{1 + (y')^2}

We are given that Length(PYP)=1\text{Length}(P Y_P) = 1 for every point on the curve. Therefore: x1+(y′)2=1x \sqrt{1 + (y')^2} = 1

Squaring both sides and simplifying: 1+(y′)2=1x21 + (y')^2 = \frac{1}{x^2} (y′)2=1−x2x2(y')^2 = \frac{1 - x^2}{x^2}

Since the curve lies in the first quadrant and passes through (1,0)(1, 0), as xx decreases from 11 towards 00, yy must increase to remain positive (y>0y > 0). Thus, the derivative y′=dydxy' = \frac{dy}{dx} must be negative for x∈(0,1)x \in (0, 1): y′=−1−x2xy' = -\frac{\sqrt{1 - x^2}}{x}

Rearranging this differential equation gives: xy′+1−x2=0x y' + \sqrt{1 - x^2} = 0

This matches Option (B).

Now, we solve the differential equation to find y(x)y(x): y=−∫1−x2x dxy = -\int \frac{\sqrt{1 - x^2}}{x} \, dx

Let x=sin⁡θx = \sin\theta, which gives dx=cos⁡θ dθdx = \cos\theta \, d\theta for θ∈(0,π2]\theta \in \left(0, \frac{\pi}{2}\right]: y=−∫1−sin⁡2θsin⁡θ⋅cos⁡θ dθ=−∫cos⁡2θsin⁡θ dθy = -\int \frac{\sqrt{1 - \sin^2\theta}}{\sin\theta} \cdot \cos\theta \, d\theta = -\int \frac{\cos^2\theta}{\sin\theta} \, d\theta y=∫sin⁡2θ−1sin⁡θ dθ=∫(sin⁡θ−csc⁡θ) dθy = \int \frac{\sin^2\theta - 1}{\sin\theta} \, d\theta = \int (\sin\theta - \csc\theta) \, d\theta y=−cos⁡θ+log⁡e∣csc⁡θ+cot⁡θ∣+Cy = -\cos\theta + \log_e|\csc\theta + \cot\theta| + C

Substituting back sin⁡θ=x\sin\theta = x, cos⁡θ=1−x2\cos\theta = \sqrt{1 - x^2}, csc⁡θ=1x\csc\theta = \frac{1}{x}, and cot⁡θ=1−x2x\cot\theta = \frac{\sqrt{1 - x^2}}{x}: y=log⁡e(1+1−x2x)−1−x2+Cy = \log_e \left( \frac{1 + \sqrt{1 - x^2}}{x} \right) - \sqrt{1 - x^2} + C

Using the initial condition that the curve passes through (1,0)(1, 0), i.e., y(1)=0y(1) = 0: 0=log⁡e(1)−0+C  ⟹  C=00 = \log_e(1) - 0 + C \implies C = 0

Thus, the explicit equation for the curve is: y=log⁡e(1+1−x2x)−1−x2y = \log_e \left( \frac{1 + \sqrt{1 - x^2}}{x} \right) - \sqrt{1 - x^2}

This matches Option (A).

Correct Options: (A) and (B)

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