Let $\Gamma$ denote a curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ 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 $\mathrm{Y}_{\mathrm{P}}$. If $\mathrm{PY}_{\mathrm{P}}$ has length 1 for each point P on $\Gamma$, then which of the following options is/are correct ?

(A) $\mathrm{y}=\log _{\mathrm{e}}\left(\frac{1+\sqrt{1-\mathrm{x}^{2}}}{\mathrm{x}}\right)-\sqrt{1-\mathrm{x}^{2}}$; (D) $x y^{\prime}+\sqrt{1-x^{2}}=0$

(B) $\mathrm{xy}^{\prime}-\sqrt{1-\mathrm{x}^{2}}=0$

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

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Differential Equations

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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) $y = lo g_{e} (\frac{1 + 1 - x ^{2}}{x}) - 1 - x^{2}$ ✓ Correct
(B) $xy^{'} - 1 - x^{2} = 0$
(C) $y = - lo g_{e} (\frac{1 + 1 - x ^{2}}{x}) + 1 - x^{2}$
(D) $x y^{'} + 1 - x^{2} = 0$ ✓ Correct

Answer: (A), (D)

Solution

The tangent at

P (x, y)

Y - y = y^{'} (X - x)

.

At the y-axis (

X = 0

Y_{P} = (0, y - xy^{'})

.

The distance

PY_{P} = 1

gives:

x^{2} + (xy^{'})^{2} = 1

Since the curve is in the first quadrant and passes through

(1, 0)

y

must be decreasing, so:

\frac{dy}{dx} = - \frac{1 - x ^{2}}{x ^{2}}

This gives

x y^{'} + 1 - x^{2} = 0

✓

Integrating with substitution

x = sin θ

y = ln (\frac{1 + 1 - x ^{2}}{x}) - 1 - x^{2}

Using the initial condition

y (1) = 0

confirms this solution. ✓

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Differential Equations

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