The equation of the parabola having focus at $(-1,-2)$ and directrix is $x-2 y+3=0$, is ........... .
Given that, focus at $F(-1,-2)$ and directrix is $x-2 y+3=0$
Let any point on the parabola be $(x, y)$.
$$\begin{array}{ll} \therefore & \quad P F=\left|\frac{x-2 y+3}{\sqrt{1+4}}\right| \\ \Rightarrow & (x+1)^2+(y+2)^2=\frac{(x-2 y+3)^2}{5} \\ \Rightarrow & 5\left[x^2+2 x+1+y^2+4 y+4\right]=x^2+4 y^2+9-4 x y-12 y+6 x \\ \Rightarrow & 4 x^2+y^2+4 x+32 y+16=0 \end{array}$$
The equation of the hyperbola with vertices at $(0, \pm 6)$ and eccentricity $\frac{5}{3}$ is ........... and its foci are ........... .
Let the equation of the hyperbola be $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Then vertices $=(0, \pm b)=(0, \pm 6)$
$$\begin{array}{llrl} \therefore & b =6 \text { and } e=5 / 3 \\ \because & e =\sqrt{1+\frac{a^2}{b^2}} \Rightarrow \frac{25}{9}=1+\frac{a^2}{36} \\ \Rightarrow & \frac{25-9}{9} =\frac{a^2}{36} \Rightarrow 16=\frac{a^2}{4} \Rightarrow a^2=48 \end{array}$$
So, the equation of hyperbola is,
$$\begin{aligned} \frac{-x^2}{48}+\frac{y^2}{36} & =1 \Rightarrow \frac{y^2}{36}-\frac{x^2}{48}=1 \\ \text { Foci } & =(0, \pm \text { be })=\left(=0, \pm \frac{5}{3} \times 6\right)=(0, \pm 10) \end{aligned}$$
The area of the circle centred at $(1,2)$ and passing through the point $(4,6)$ is
Equation of a circle which passes through $(3,6)$ and touches the axes is
Equation of the circle with centre on the $Y$-axis and passing through the origin and the point $(2,3)$ is