Question

Let y = f(x) be a curve $C_1$ passing through the points (1, 1) and $(4,\frac{1}{4})$and satisfying a differential equation $y\left(\frac{d^{2}y}{d{x}^2}\right)=2{\left(\frac{dy}{dx}\right)}^2$. Curve $C_2$ is the director circle of the circle $x^2+y^2=1$, then the shortest distance between curves $C_1$ and $C_2$ is

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

The correct option is B 0

Given differential equation is
$yy"=2(y^{\prime }{)}^2$
$\Rightarrow \int \frac{y"}{y^{\prime }}=\int \frac{2}{y}{y}^{\prime }$
$\Rightarrow \log y^{\prime }=2\log y+\log a$
$\Rightarrow \log y^{\prime }=\log ya{y}^2$
$\Rightarrow \int \frac{y^{\prime }}{y^2}=a\int dx$
$\Rightarrow -\frac{1}{y}ax+b$
But curve is passing through the points (1, 1) and $(4,\frac{1}{4})$
$\therefore -1=a+b.....\left(i\right)$
and $-4=4a+b.....\left(ii\right)$

$\therefore$ From (i) -1 = -1 + b
$\Rightarrow b=0$
$\therefore -\frac{1}{y}=(-1)x+0\Rightarrow xy=1$
$\therefore$ Curve $C_1:xy=1......\left(i\right)$
and equation of director circle w.r.t. circle $x^2+y^2=1$ is $x^2+y^2=(\sqrt{2}.1{)}^2=2$
$\therefore$ Cureve $C_2:x^2+y^2=2....\left(ii\right)$
$\therefore$ Shortest distance between curves $C_1$ and $C_2=$ distance of the point (1,1) in curve $C_1$ - radius of curve $C_2$ from origin
$=\sqrt{(1-0{)}^2+(1-0{)}^2}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0$