⇒Defferential Equation

Correct Option is (A)

Circle passes through origin and centre lie on Y-axis.

Let $(0,k)$ be centre and '$k$' be radius

$\therefore$ Equation of circle is

$\begin{array}{rl}& (x-0{)}^2+(y-\mathrm{k}{)}^2={\mathrm{k}}^2\\ & x^2+y^2-2y\mathrm{k}+{\mathrm{k}}^2={\mathrm{k}}^2\\ & x^2+y^2-2\mathrm{k}y=0\\ & x^2+y^2=2\mathrm{k}y\text{.... (i)}\\ & \frac{x^2+y^2}{2y}=\mathrm{k}\text{.... (ii)}\end{array}$

Differentiating equation (i) with respect to $x$, we get

$\begin{array}{rl}& 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=2\mathrm{k}\frac{\mathrm{d}y}{\mathrm{d}x}\\ & 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}-2\mathrm{k}\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & 2x+2(y-\mathrm{k})\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & 2x+2[y-\left(\frac{x^2+y^2}{2y}\right)]\frac{\mathrm{d}y}{\mathrm{d}x}=0\text{...[From (ii)]}\\ & 2x+2\left[\frac{2y^2-x^2-y^2}{2y}\right]\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & 2x+\left(\frac{y^2-x^2}{y}\right)\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & 2xy+(y^2-x^2)\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & \text{ i.e. }(x^2-y^2)\frac{\mathrm{d}y}{\mathrm{d}x}-2xy=0\end{array}$

Correct Option is (C)

Parabola whose axes are parallel to $\mathrm{Y}$-axis. Vertex is not $(0,0)$

Equation becomes

$(x-\mathrm{h}{)}^2=4\mathrm{b}(y-\mathrm{k})$

Differentiating w.r.t. $x$, we get

$2(x-h)=4b\left(\frac{dy}{dx}\right)$

Again differentiating w.r.t. $x$, we get

$2=4b\left(\frac{d^{2}y}{d{x}^2}\right)$

Again differentiating w.r.t. $x$, we get

$\begin{array}{r}0=4b\left(\frac{d^{3}y}{d{x}^3}\right)\\ \therefore \frac{d^{3}y}{d{x}^3}=0\\ \text{ i.e., }{y}_3=0\end{array}$

Correct Option is (B)

Since circles touch $\mathrm{Y}$ axis at origin, the centres of the circles lie on $\mathrm{X}$ axis.

Let centre be $(h,0)$ and radius $=h$.

$\therefore (\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-0{)}^2={\mathrm{h}}^2\Rightarrow {\mathrm{x}}^2-2\mathrm{hx}+{\mathrm{y}}^2=0$ ..... (1)

Differentiating w.r.t. $x$, we get

$2x-2h+2y\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}=h$

Substituting value of $h$ in eq. (1), we get

$\begin{array}{rl}& x^2-2(x+y\frac{dy}{dx})x+y^2=0\\ \therefore & x^2-2x^2-2xy\frac{dy}{dx}+y^2=0\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0\end{array}$

Correct Option is (B)

Let $(\mathrm{h},0)$ be the centre of the circle and '$r$' be the radius.

Let $(\mathrm{h},\mathrm{h}-\mathrm{h}{)}^2+(\mathrm{y}-0{)}^2={\mathrm{r}}^2\Rightarrow (\mathrm{x}-\mathrm{h}{)}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ ....... (1)

Differentiating w.r.t. $x$, we get

$\therefore 2(x-h)(1)+2y\frac{dy}{dx}=0\Rightarrow h=x+y\frac{dy}{dx}$

Substituting value of '$h$' in eq. (i), we get

$y^2{\left(\frac{dy}{dx}\right)}^2+y^2=r^2\Rightarrow y^2[1+{\left(\frac{dy}{dx}\right)}^2]=r^2$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& y^2(2\frac{dy}{dx}\cdot \frac{d^{2}y}{d{x}^2})+[1+{\left(\frac{dy}{dx}\right)}^2](2y)\frac{dy}{dx}=0\\ & \therefore y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2+1=0\end{array}$

Correct Option is (A)

For the given ellipse, we have $a=2b$

$\therefore \frac{x^2}{4b^2}+\frac{y^2}{b^2}=1\Rightarrow x^2+4y^2=4b^2$

Differentiating both sides w.r.t. $x$, we get

$2x+8y\frac{dy}{dx}=0\Rightarrow x+4y\frac{dy}{dx}=0$