Let \Gamma denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to \Gamma at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on \Gamma, then which of the following options is/are correct?

Engineering

Mathematics

Variable Separable Differential Equation

Question

Let $Γ$ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to $Γ$ at a point P intersect the y-axis at Y_P. If PY_P has length 1 for each point P on $Γ$ , then which of the following options is/are correct?

A

$y = {ℓog}_{e} (\frac{1 + \sqrt{1 - x^{2}}}{x}) - \sqrt{1 - x^{2}}$

B

$xy' + \sqrt{1 - x^{2}} = 0$

C

$xy' - \sqrt{1 - x^{2}} = 0$

D

$y = - {ℓog}_{e} (\frac{1 + \sqrt{1 - x^{2}}}{x}) + \sqrt{1 - x^{2}}$

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Y – y = y' (X – x)

So, Y_P = (0, y – xy')

So, x² + (xy')² = 1 $\Rightarrow \frac{dy}{dx} = - \sqrt{\frac{1 - x^{2}}{x^{2}}}$

[ $\frac{dy}{dx}$ cannot be positive i.e. f(x) cannot be increasing in first quadrant for x $\in$ (0, 1)]

Hence, $\int dy = - \int \frac{\sqrt{1 - x^{2}}}{x} dx$

⇒ y = – $\int \frac{\cos^{2} θ}{sinθ} dθ$ ; put x = sin θ

⇒ y = – $\int$ cosec θ dθ + $\int$ sin θ dθ

⇒ y = ln (cosec θ + cot θ) – cos θ + C

⇒ y = $ℓn (\frac{1 + \sqrt{1 - x^{2}}}{x}) - \sqrt{1 - x^{2}} + C$

⇒ y = $ℓn (\frac{1 + \sqrt{1 - x^{2}}}{x}) - \sqrt{1 - x^{2}}$ [As y(A) = 0]