If $c$ is a parameter, then the differential equation of the family of curves $x^2=c(y+c)^2$ is

Question

If $c$ is a parameter, then the differential equation of the family of curves $x^2=c(y+c)^2$ is, $x\left(\frac{d y}{d x}\right)^3+y\left(\frac{d y}{d x}\right)^2-1=0$, $x\left(\frac{d y}{d x}\right)^3-y\left(\frac{d y}{d x}\right)^2+1=0$, $x\left(\frac{d y}{d x}\right)^3+y\left(\frac{d y}{d x}\right)^2+1=0$, $x\left(\frac{d y}{d x}\right)^3-y\left(\frac{d y}{d x}\right)^2-1=0$

MARKS App · Accepted Answer

Given equation of the family of curves $\begin{aligned} & x^2=c(y+c)^2 \ & x=\sqrt{c}(y+c) \quad \ldots (i) \end{aligned}$ [on taking square root both sides] Now, differentiate both sides w.r.t. $x$, we get $\begin{aligned} \mathrm{I} & =\sqrt{c} \frac{d y}{d x} \ \Rightarrow \quad \sqrt{c} & =\frac{d x}{d y} \quad \ldots (ii) \end{aligned}$ From Eqs. (i) and (ii) on eliminating ' $c$ ', we get $\begin{aligned} & x=\left(\frac{d x}{d y}\right)\left[y+\left(\frac{d x}{d y}\right)^2\right] \ & \Rightarrow \quad x\left(\frac{d y}{d x}\right)^3=y\left(\frac{d y}{d x}\right)^2+1 \ & \Rightarrow \quad x\left(\frac{d y}{d x}\right)^3-y\left(\frac{d y}{d x}\right)^2-1=0 \ \end{aligned}$ Hence, option (4) is correct.

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