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

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

$\frac{-1+h}{2}=1and\frac{1+k}{2}=3\phantom{\rule{0ex}{0ex}}h=3andk=5$

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

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

${m}_{1}=slopeofAS=\frac{1-3}{-1-1}=\frac{-2}{-2}=1$

let ${m}_{2}$ be the slope of the directrix. then,

${m}_{1}.{m}_{2}=-1\phantom{\rule{0ex}{0ex}}{m}_{2}=-\frac{1}{{m}_{1}}=-1$

thus the directrix passes through the point (3,5) and has slope -1.

the equation of the directrix is

$y-5=-1(x-3)\phantom{\rule{0ex}{0ex}}y+x=3+5\phantom{\rule{0ex}{0ex}}x+y=8.......\left(1\right)$

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,

$SP=PM\Rightarrow S{P}^{2}=P{M}^{2}$

$\Rightarrow (x+1{)}^{2}+(y-1{)}^{2}={\left(\frac{x+y-8}{\sqrt{1+1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}+1+2x+{y}^{2}-2y+1=\frac{(x+y-8{)}^{2}}{2}\phantom{\rule{0ex}{0ex}}2*({x}^{2}+{y}^{2}+2x-2y+2)={x}^{2}+{y}^{2}+{8}^{2}+2xy-16x-16y\phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}+4x+16x-4y+16y+4=64+2xy\phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}-2xy+20x+12y-60=0$

this is the equation of the required parabola.

hope this helps you

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