Q88 / 160 · TS EAMCET 2003 · Mathematics
The differential equation of the family of parabola with focus as the origin and the axis as $X$-axis, is :
- A$y\left(\frac{d y}{d x}\right)^2+4 x \frac{d y}{d x}=4 y$
- Correct answerB$-y\left(\frac{d y}{d x}\right)^2=2 x \frac{d y}{d x}-y$
- C$y\left(\frac{d y}{d x}\right)^2+y=2 x y \frac{d y}{d x}$
- D$y\left(\frac{d y}{d x}\right)^2+2 x y \frac{d y}{d x}+y=0$
Solution
Given that,
Focus $S=(0,0)$, let $P(x, y)$ be any point on the parabola.
Since, $S P^2=P M^2$
$\Rightarrow \quad(x-0)^2+(y-0)^2=(x+a)^2$
$\Rightarrow \quad x^2+y^2=x^2+a^2+2 x a$
$\Rightarrow \quad y^2=2 a x+a^2$ $\ldots$ (i)
$\Rightarrow \quad 2 y \frac{d y}{d x}=2 a$ $\ldots$ (ii)
From Eqs. (i) and (ii), we get
$y^2=2 y \frac{d y}{d x} \cdot x+\left(y \frac{d y}{d x}\right)^2$
$\Rightarrow \quad y^2\left(\frac{d y}{d x}\right)^2+2 x y \frac{d y}{d x}=y^2$
$\Rightarrow \quad y\left(\frac{d y}{d x}\right)^2+2 x \frac{d y}{d x}=y$
$\therefore \quad-y\left(\frac{d y}{d x}\right)^2=2 x \frac{d y}{d x}-y$
Hence, the correct answer is option: 'B'.