Find the equation of a parabola whose vertex is (1,3) and focus is(-1,1)

Solution :

In a parabola , vertex is the mid-point of the focus and the point of intersection of the axis of parabola and directrix.
Let (h,k) be the coordinates of the point of intersection of the axis and directrix.
Then (1,3) is the mid-point of the line segment joining (-1,1) and (h,k).
therefore by mid - point formula :

$$\begin{gathered} \frac{-1+\mathrm{h}}{2}=1 \mathrm{and} \frac{1+\mathrm{k}}{2}=3 \\ \Rightarrow \mathrm{h}=3 \mathrm{and} \mathrm{k}=5 \end{gathered}$$

Thus directrix meets the axis at (3,5).

Now, slope of line joining points $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{and} \left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$  = $\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

Let A be the vertex and S be the focus of the required parabola. then,

${\mathrm{m}}_1=\mathrm{slope} \mathrm{of} \mathrm{AS}=\frac{1-3}{-1-1}=\frac{-2}{-2}=1$         
Let $m_2$ be the slope of the directrix. then,

$$\begin{gathered} {\mathrm{m}}_1 {\mathrm{m}}_2=-1 \mathbf{ }\left[\mathrm{as} \mathrm{product} \mathrm{of} \mathrm{the} \mathrm{slopes} \mathrm{of} \mathrm{two} \mathrm{mutually} \mathrm{perpendicular} \mathrm{lines} \mathrm{is} -1\right] \\ \Rightarrow {\mathrm{m}}_2=-\frac{1}{{\mathrm{m}}_1}=-1 \end{gathered}$$
Thus the directrix passes through the point (3,5) and has slope -1.

We know that equation of a line passing through the point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and having slope "m" is

$\mathrm{y} - {\mathrm{y}}_1 = \mathrm{m}\left(\mathrm{x} - {\mathrm{x}}_1\right)$

The equation of the directrix is:

$$\begin{gathered} \mathrm{y}-5=-1(\mathrm{x}-3) \\ \Rightarrow \mathrm{y}+\mathrm{x}=3+5 \\ \Rightarrow \mathrm{x}+\mathrm{y}=8 .......\left(1\right) \end{gathered}$$

Let P(x,y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix.
then,
$\mathrm{SP}=\mathrm{PM}\Rightarrow {\mathrm{SP}}^2={\mathrm{PM}}^2$
$$\begin{gathered} \Rightarrow (\mathrm{x}+1{)}^2+(\mathrm{y}-1{)}^2={\left(\frac{\mathrm{x}+\mathrm{y}-8}{\sqrt{1+1}}\right)}^2 \\ \Rightarrow {\mathrm{x}}^2+1+2\mathrm{x}+{\mathrm{y}}^2-2\mathrm{y}+1=\frac{(\mathrm{x}+\mathrm{y}-8{)}^2}{2} \\ \Rightarrow 2({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}+2)={\mathrm{x}}^2+{\mathrm{y}}^2+8^2+2\mathrm{xy}-16\mathrm{x}-16\mathrm{y} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+16\mathrm{x}-4\mathrm{y}+16\mathrm{y}+4=64+2\mathrm{xy} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}+20\mathrm{x}+12\mathrm{y}-60=0 \end{gathered}$$

This is the equation of the required parabola.