The differential equation obtained from the function $y=a(x-a)^{2}$ is
The differential equation obtained from the function $y=a(x-a)^{2}$ is
- $8 y^{2}=\left(\frac{d y}{d x}\right)^{2}\left[x-\frac{1}{4 y}\left(\frac{d y}{d x}\right)^{2}\right]^{2}$
- $8 y^{2}=\left(\frac{d y}{d x}\right)^{2}\left[x+\frac{1}{4 y}\left(\frac{d y}{d x}\right)^{2}\right]^{2}$
- $2 y^{2}=\left(\frac{d y}{d x}\right)^{2}\left[x-\frac{1}{4 y}\left(\frac{d y}{d x}\right)^{2}\right]^{2}$
- $4 y^{2}=\left(\frac{d y}{d x}\right)^{2}\left[x-\frac{1}{4 y}\left(\frac{d y}{d x}\right)^{2}\right]^{2}$
Solution
$\begin{aligned}
y &=a(x-a)^{2}...(1) \\
\therefore \frac{d y}{d x} &=2 a(x-a)...(2)
\end{aligned}$
Eq. $(2)+\mathrm{Eq} \cdot(1)$ gives
$\frac{\left(\frac{d y}{d x}\right)}{y}=\frac{2 a(x-a)}{a(x-a)^{2}} \quad \Rightarrow(x-a)=\frac{2 y}{\left(\frac{d y}{d x}\right)} \text { and } a=x-\frac{2 y}{\left(\frac{d y}{d x}\right)}$
Substituting these values of $a$ and $(x-a)$ in eq. (1), we get
$\begin{aligned}
y &=\left[x-\frac{2 y}{\left(\frac{d y}{d x}\right)}\right]\left[\frac{2 y}{\left(\frac{d y}{d x}\right)}\right]^{2} \\
&=\left[\frac{x\left(\frac{d y}{d x}\right)-2 y}{\left(\frac{d y}{d x}\right)}\right]\left[\frac{4 y^{2}}{\left(\frac{d y}{d x}\right)^{2}}\right] \\
\therefore & \frac{(y)\left(\frac{d y}{d x}\right)^{3}}{x\left(\frac{d y}{d x}-2 y\right)}=4 y^{2}
\end{aligned}$
Asked in: MHT CET 2020 (16 Oct Shift 1)
Practice more Hyperbola questions on Aicharya