Question 891029
Using the value of the latus rectum,
{{{y-k=(1/6)(x-h)^2}}}
{{{6(y-k)=(x-h)^2}}}
Now use both points,
{{{6(-3-k)=(-1-h)^2}}}
{{{-18-6k=-(h^2+2h+1)}}}
{{{h^2+2h+1+6k=-18}}}
1.{{{h^2+2h+6k=-19}}}
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{{{6(3/2-k)=(2-h)^2}}}
{{{9-6k=4-4h+h^2}}}
2.{{{h^2-4h+6k=5}}}
Subtracting eq. 2 from eq. 1,
{{{h^2+2h+6k-h^2+4h-6k=-19-5}}}
{{{6h=-24}}}
{{{h=-4}}}
Then,
{{{(-4)^2+2(-4)+6k=-19}}}
{{{16-8+6k=-19}}}
{{{6k=-27}}}
{{{k=-9/2}}}
So then the equation is,
{{{y+9/2=(1/6)(x+4)^2}}}
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{{{drawing(300,300,-10,10,-10,10,grid(1),circle(-1,-3,0.4),circle(2,3/2,0.4),graph(300,300,-10,10,-10,10,(1/6)(x+4)^2-9/2))}}}