JEE Advanced · Mathematics · 29. Differential Eqns

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 which of the following is options is/are correct?

A $y = {l o g}_{e} (\frac{1 + \sqrt{1 - x^{2}}}{x}) - \sqrt{1 - x^{2}}$
B $x y^{'} - \sqrt{1 - x^{2}} = 0$
C $y = - {l o g}_{e} (\frac{1 + \sqrt{1 - x^{2}}}{x}) + \sqrt{1 - x^{2}}$
D $x y^{'} + \sqrt{1 - x^{2}} = 0$

Verified Solution

Answer & Solution

Correct Answer

(D) $x y^{'} + \sqrt{1 - x^{2}} = 0$

Step-by-step Solution

Detailed explanation

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} (0, k - h m_{T})

Now, as given

P Y_{P} = 1

\sqrt{h^{2} + h^{2} m_{T}^{2}} = 1

\sqrt{h^{2} (1 + m_{T}^{2})} = 1

1 + m_{T}^{2} = \frac{1}{h^{2}}

m_{T}^{2} = \frac{1}{h^{2}} - 1

m_{T}^{2} = \frac{1 - h^{2}}{h^{2}}

m_{T} = \pm \frac{\sqrt{1 - h^{2}}}{h}

Put

m_{T} = \frac{d y}{d x} a n d h = x

\frac{d y}{d x} = \pm \frac{\sqrt{1 - x^{2}}}{x}

...(i)

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

As the curve lies in

1^{s t}

quadrant y must be positive. Hence,

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

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

\frac{d y}{d x} = - \frac{\sqrt{1 - x^{2}}}{x}

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, where

a > b > 0

, be a hyperbola in the xy-plane whose conjugate axis

L M

subtends an angle of

60^{o}

at one of its vertices

N

. Let the area of the triangle

L M N

4 \sqrt{3}

LIST-I	LIST-II
A. The length of the conjugate axis of $H$ is	P. $8$
B. The eccentricity of $H$ is	Q. $\frac{4}{\sqrt{3}}$
C. The distance between the foci of $H$ is	R. $\frac{2}{\sqrt{3}}$
D. The length of the latus rectum of $H$ is	S. $4$

The correct option is:

JEE Advanced 2018 Medium

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Let T denote a curve y=yx 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 YP. If PYP has length 1 for each point P on T, then which of the following is options is/are correct?

Step-by-step Solution

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 which of the following is options is/are correct?