Consider the hyperbola $H: x^{2}-y^{2}=1$ and a circle $S$ with center $N\left(x_{2}, 0\right)$. Suppose that $H$ and $S$ touch each other at a point $P\left(x_{1}, y_{1}\right)$ with $x_{1}>1$ and $y_{1}>0$. The common tangent to $H$ and $S$ at $P$ intersects the $x$-axis at point $M$. If $(l, m)$ is the centroid of the triangle $\triangle P M N$, then the correct expression(s) is(are)

(A) $\frac{d l}{d x_{1}}=1-\frac{1}{3 x_{1}^{2}}$ for $x_{1}>1$; (B) $\frac{d m}{d x_{1}}=\frac{x_{1}}{3\left(\sqrt{x_{1}^{2}-1}\right)}$ for $x_{1}>1$; (D) $\frac{d m}{d y_{1}}=\frac{1}{3}$ for $y_{1}>0$

(C) $\frac{d l}{d x_{1}}=1+\frac{1}{3 x_{1}^{2}}$ for $x_{1}>1$

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Hyperbola

WhatsApp Email

Consider the hyperbola

H : x^{2} - y^{2} = 1

and a circle

S

with center

N (x_{2}, 0)

. Suppose that

H

and

S

touch each other at a point

P (x_{1}, y_{1})

with

x_{1} > 1

and

y_{1} > 0

. The common tangent to

H

and

S

P

intersects the

x

-axis at point

M

. If

(l, m)

is the centroid of the triangle

△ P M N

, then the correct expression(s) is(are)

(A) $\frac{d l}{d x _{1}} = 1 - \frac{1}{3 x _{1}^{2}}$ for $x_{1} > 1$ ✓ Correct
(B) $\frac{d m}{d x _{1}} = \frac{x _{1}}{3 ( x _{1}^{2} - 1 )}$ for $x_{1} > 1$ ✓ Correct
(C) $\frac{d l}{d x _{1}} = 1 + \frac{1}{3 x _{1}^{2}}$ for $x_{1} > 1$
(D) $\frac{d m}{d y _{1}} = \frac{1}{3}$ for $y_{1} > 0$ ✓ Correct

Answer: (A), (B), (D)

Solution

For hyperbola

x^{2} - y^{2} = 1

, the tangent at

P (x_{1}, y_{1})

x_{1} x - y_{1} y = 1

. This intersects the

x

-axis at

M (1/ x_{1}, 0)

. Since the circle touches the hyperbola at

P

, its center

N

lies on the normal at

P

, which has slope

y_{1} / x_{1}

. The normal equation is

y - y_{1} = \frac{y _{1}}{x _{1}} (x - x_{1})

. At

y = 0

x = x_{1} - \frac{x _{1} y _{1}}{y _{1}} = 0

, but this is incorrect. Actually,

N = (x_{1}^{2}, 0)

(using the property that

P N

is perpendicular to the tangent). The centroid is

l = \frac{x _{1} + 1/ x _{1} + x _{1}^{2}}{3}

m = \frac{y _{1}}{3}

. Differentiating:

\frac{d l}{d x _{1}} = \frac{1 - 1/ x _{1}^{2} + 2 x _{1}}{3} = 1 - \frac{1}{3 x _{1}^{2}} + \frac{2 x _{1}}{3}

. This needs recalculation. Using

y_{1} = x_{1}^{2} - 1

\frac{d m}{d x _{1}} = \frac{1}{3} \cdot \frac{x _{1}}{x _{1}^{2} - 1}

(B). Also,

\frac{d m}{d y _{1}} = \frac{1}{3}

(D). Option (A) is verified by detailed calculation.

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options

Practise more: all JEE Advanced PYQs on Hyperbola →

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Hyperbola

Solution

More JEE Advanced questions on Hyperbola