# The focus of parabola is (1,5) and its directrix is x + y + 2 = 0. Find the equation of the parabola, its vertex and length of latus rectum.

let F (1,5) be the focus of the parabola and the equation of the directrix is $x+y+2=0$...........(1)

let P(h.k) be any point on the parabola.
the distance from the focus to the point P = the distance from the directrix to the point P
$PM=PF$
$\sqrt{\left(h-1{\right)}^{2}+\left(k-5{\right)}^{2}}=\left|\frac{h*1+k*1+2}{\sqrt{1+1}}\right|\phantom{\rule{0ex}{0ex}}\left(h-1{\right)}^{2}+\left(k-5{\right)}^{2}=\frac{\left(h+k+2{\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}2*\left[{h}^{2}+1-2h+{k}^{2}+25-10k\right]={h}^{2}+{k}^{2}+4+2hk+4k+4h\phantom{\rule{0ex}{0ex}}2{h}^{2}+2{k}^{2}-4h-20k+52={h}^{2}+{k}^{2}+4+2hk+4k+4h\phantom{\rule{0ex}{0ex}}{h}^{2}+{k}^{2}-8h-24k-2hk+48=0$
put h = x and k = y
${x}^{2}+{y}^{2}-8x-24y-2xy+48=0$ is the required equation of the parabola.

the axis of parabola passes through the focus and perpendicular to the directrix.
the slope of axis = $-\left(-1\right)=1$ [since slope of directrix is -1]
the equation of axis is

the point of intersection of axis and directrix is intersection point of eq(1) and eq(2)
$x+y+2=0\phantom{\rule{0ex}{0ex}}x-y+4=0\phantom{\rule{0ex}{0ex}}2x+6=0⇒x=-3\phantom{\rule{0ex}{0ex}}-3+y+2=0\phantom{\rule{0ex}{0ex}}⇒y=1$
the coordinates of k are (-3,1).
vertex is the mid-point of focus and point K
i.e. coordinates of vertex are
now distance from vertex to focus = $\sqrt{\left(-1-1{\right)}^{2}+\left(3-5{\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
the length of the latusrectum = $4*2\sqrt{2}=8\sqrt{2}$

hope this helps you

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