Find the equation of the parabola in which the ends of the latus rectum have the coordinates $(-1,5)$ and $(-1,-11)$ and the vertex is $(-5,-3)$.
I could think of assuming the equation of parabola as $(y-k)^2=4a(x-h)$ and plug in those three points to get three linear equations and solve for the unknowns. But, I wonder if there's any other approach?
$\endgroup$2 Answers
$\begingroup$$h$ and $k$ are the coordinates of the vertex, hence $h=-5$ and $k=-3$. As focus is $F=(-1,-3)$ (midpoint of latus rectum) and vertex $V=(-5,-3)$, then $a=VF=4$.
$\endgroup$ $\begingroup$Vertex $(h,k) = (-5,-3)$.
We know, Length of Latus rectum $=4a$, and the end points are $(-1,5)$ and $(-1,-11)$.
Now using the distance formula, we have$$4a=[ (-1+1)^2 + (-11-5 )^2 ] ^{1/2}$$
or,$$4a =(0+256)^{1/2}$$
or $4a=16$. Therefore $a=4$.
So the required equation of parabola is $$(y-k)^2 = 4a(x-h)$$
or,$$(y+3)^2= 4\cdot4(x+5)$$
or,$$y^2+ 6y +9 = 16x + 80$$
or$$y^2 + 6y - 16x - 71 = 0$$
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