Question

Let $y=f\left(x\right)$ be a curve $C_1$ passing through the points (1, 1) and $(4,\frac{1}{4})$ and satisfying differential equation $y\left(\frac{d^{2}y}{d{x}^2}\right)=2(\frac{dy}{dx}{)}^2$. 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. 0
• B. 1
• C. 3
• D. 2

Answer 1

The correct option is B 1

Given differential equation is $y{y}^{\prime \prime }=2(y^{\prime }{)}^2$
$\Rightarrow \int \frac{y^{\prime \prime }}{y^{\prime }}=\int \frac{2}{y}{y}^{\prime }$
$\Rightarrow log{y}^{\prime }=2logy+loga$
$\Rightarrow log{y}^{\prime }=logya{y}^2$
$\Rightarrow \int \frac{y^{\prime }}{y^2}=a\int dx$

$\Rightarrow -\frac{1}{y}ax+b$
But curve is passing through the points (1, 1) and $(4,\frac{1}{4})$
$\therefore -1=a+b$ ...(i)
and $-4=4a+b$ ... (ii)
By [(ii) - (i)]
----------------
$4a+b=-4$
$a+b=-1$
- - +
------------------
$3a=-3\Rightarrow a=-1$
$From\left(i\right)-1=-1+b$
$\Rightarrow b=0$

$\therefore -\frac{1}{y}=(-1).x+0\Rightarrow xy=1$
$\therefore CurveC:xy=1$.... (i)
and equation of durector circle w.r.t circle $x^2+y^2=1$ is $x^2+y^2=(\sqrt{2}.1{)}^2=2$

$\therefore CurveC:x^2+y^2=2$ .... (ii)
$\therefore$ 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

$=\sqrt{(1-0{)}^2+(1-0{)}^2}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0$