Let $y = f (x)$ be a curve $C_{1}$ passing through $(2, 2)$ and $(8, \frac{1}{2})$ and satisfying a differential equation $y (\frac{d^{2} y}{d x^{2}}) = 2 {(\frac{d y}{d x})}^{2}$ . Curve $C_{2}$ is the director circle of the circle $x^{2} + y^{2} = 2$ . If the shortest distance between the curves $C_{1}$ and $C_{2}$ is $\sqrt{p} - q$ where $p, q \in N$ , then the value of ( $p^{2} - q$ ) is

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Solution

Given
$y (\frac{d^{2} y}{d x^{2}}) = 2 {(\frac{d y}{d x})}^{2}$
$\Rightarrow \frac{y^{''}}{y^{'}} = \frac{2 y^{'}}{y} \Rightarrow ln y^{'} = 2 ln y + ln a$
$\Rightarrow \frac{y^{'}}{y^{2}} = a \Rightarrow \int \frac{d y}{y^{2}} = \int a d x$
$\Rightarrow - \frac{1}{y} = a x + b$
Since curve is passing through $(2, 2)$ and $(8, \frac{1}{2})$ , so

$2 a + b = - \frac{1}{2} \dots (1)$
$8 a + b = - 2 \dots (2)$
$∴$ On solving (1) and (2), we get $a = - \frac{1}{4}, b = 0$
Hence $C_{1} : x y = 4$ and curve $C_{2}$ is $x^{2} + y^{2} = 4$ .
$∴$ Shortest distance between $C_{1}$ and $C_{2}$ is $= 2 \sqrt{2} - 2 = \sqrt{8} - 2$
Hence $p^{2} - q = 64 - 2$

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