JEE Advanced 2016 Paper 2 · Mathematics · Differential Equations · hard

The curve y=y(x), x>1, satisfies the differential equation 2xy dy/dx=x^2+y^2 and passes through (4, 2√3).

One or more options correct

The curve

y=y(x)

x>1

, satisfies the differential equation

2xy\,\dfrac{dy}{dx}=x^{2}+y^{2}

and passes through

\left(4,\,2\sqrt3\right)

. Which of the following statements is/are correct?

A. The curve is a part of a rectangular hyperbola
B. The centre of this hyperbola is $\left(\tfrac12,0\right)$
C. $y(2)=2$
D. The tangent to the curve at $\left(4,2\sqrt3\right)$ meets the $x$ -axis at $\left(\tfrac47,0\right)$

Show answer and solution

Answer: A, B, D

Put

y=vx

v+x\frac{dv}{dx}=\frac{1+v^{2}}{2v}

, so

\frac{2v\,dv}{1-v^{2}}=\frac{dx}{x}

, which integrates to

x(1-v^{2})=c

, i.e.

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

. Using

(4,2\sqrt3)

16-12=4c

gives

c=1

. So

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

, i.e.

\left(x-\frac12\right)^{2}-y^{2}=\frac14

. This is a rectangular hyperbola centred at

\left(\frac12,0\right)

, so (A) and (B) are true. At

x=2

y^{2}=4-2=2

, so

y(2)=\sqrt2\ne2

and (C) is false. The slope at

(4,2\sqrt3)

\frac{16+12}{2\cdot4\cdot2\sqrt3}=\frac{7}{4\sqrt3}

; the tangent meets

y=0

x=4-\frac{2\sqrt3}{7/(4\sqrt3)}=4-\frac{24}{7}=\frac47

, so (D) is true.

Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake