Let T 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 T at a point P intersect the y-axis at Y _ P . If P Y _ P has length 1 for each point P on T, then

Correct answer: A. y = _ e ( 1 + 1 - x ^ 2 x ) - 1 - x ^ 2 Let point P is ( h, k ) Now, equation of tangent at P ( y - k ) = m _ T ( x - h ) ...(i) ( m _ T = slope of tangent at P ) For Y _ P , put x = 0 in.....(i) Y _ P (0, k - h m _ T ) Now, as given P Y _ P = 1 h ^ 2 + h ^ 2 m _ T ^ 2 = 1 h ^ 2 (1 + m _ T ^ 2 ) = 1 1 + m _ T ^ 2 = 1 h ^ 2 m _ T ^ 2 = 1 h ^ 2 - 1 m _ T ^ 2 = 1 - h ^ 2 h ^ 2 m _ T = ± 1 - h ^ 2 h Put m _ T = (dy)/(dx)andh = x (dy)/(dx) = ± 1 - x ^ 2 x ...(i) y = ± (- ( 1 + 1 - x ^ 2 x ) + 1 - x ^ 2 ) As the curve lies in 1 ^ st quadrant y must be positive. Hence, y = ( 1 + 1 - x ^ 2 x ) - 1 - x ^ 2 Also from equation (i) only negative sign will give the correct equation of centre. Hence, (dy)/(dx) = - 1 - x ^ 2 x x y ^ ′ + 1 - x ^ 2 = 0

JEE Advanced 2019Paper 1

Exam: JEE Advanced
Year: 2019
Paper: Paper 1
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2019 · Paper 1 · Mathematics · Differential Equations

Let T denote a curve y = y (x) which is in the first quadrant and...

Let

T

denote a curve

y = y \left(x\right) which is

the first quadrant and let the point \left(1, 0\right) lie on it. Let the tangent to

at a point

intersect the y-axis at

{Y}_{P} .

If

P{Y}_{P}

has length

for each point

on

then which of the following is options is/are correct?

Let T 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 T

Options

A $y = \log_{e}\left(\dfrac{1 + \sqrt{1 - {x}^{2}}}{x}\right) - \sqrt{1 - {x}^{2}}$
B $x{y}^{′} - \sqrt{1 - {x}^{2}} = 0$
C $y = - \log_{e}\left(\dfrac{1 + \sqrt{1 - {x}^{2}}}{x}\right) + \sqrt{1 - {x}^{2}}$
D $x{y}^{′} + \sqrt{1 - {x}^{2}} = 0$

Correct answer

A. $y = \log_{e}\left(\dfrac{1 + \sqrt{1 - {x}^{2}}}{x}\right) - \sqrt{1 - {x}^{2}}$

Step-by-step solution

Let point

P

( h, k )

Now, equation of tangent at

P

( y - k ) = {m}_{T} ( x - h )

...(i) (

{m}_{T} =

slope of tangent at

P

)
For

{Y}_{P}

, put

x = 0

in.....(i)

{Y}_{P}\left(0, k - h{m}_{T}\right)

Now, as given

P{Y}_{P} = 1

\sqrt{{h}^{2} + {h}^{2}{m}_{T}^{2}} = 1

\sqrt{{h}^{2}\left(1 + {m}_{T}^{2}\right)} = 1

1 + {m}_{T}^{2} = \dfrac{1}{{h}^{2}}

{m}_{T}^{2} = \dfrac{1}{{h}^{2}} - 1

{m}_{T}^{2} = \dfrac{1 - {h}^{2}}{{h}^{2}}

{m}_{T} = \pm \dfrac{\sqrt{1 - {h}^{2}}}{h}

Put

{m}_{T} = \dfrac{dy}{dx}andh = x

\dfrac{dy}{dx} = \pm \dfrac{\sqrt{1 - {x}^{2}}}{x}

...(i)

y = \pm \left(- \ln \left(\dfrac{1 + \sqrt{1 - {x}^{2}}}{x}\right) + \sqrt{1 - {x}^{2}}\right)

As the curve lies in

{1}^{st}

quadrant y must be positive. Hence,

y = \ln \left(\dfrac{1 + \sqrt{1 - {x}^{2}}}{x}\right) - \sqrt{1 - {x}^{2}}

Also from equation (i) only negative sign will give the correct equation of centre.
Hence,

\dfrac{dy}{dx} = - \dfrac{\sqrt{1 - {x}^{2}}}{x}

x{y}^{′} + \sqrt{1 - {x}^{2}} = 0

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