# Find an equation of the parabola satisfying the following properties.

latus rectum is the line segment joining the points $(2,4)$ and $(6,4)$; passing through the point $(8,1)$.

Let $P$ be the point $(2,4)$ and $Q(6,4)$.

The distance between the $P(2,4)$ and $(6,4)$ is $4$ units.

But I don't what next step to get the equation.

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The latus rectum is the line segment through a focus of a conic section, perpendicular to the major axis, which has both endpoints on the curve.

The length of a parabola's latus rectum is $4p$, where $p$ is the distance from the focus to the vertex.

Since the length of the latus rectum of your parabola is the length of the line segment joining the points $(2, 4)$ and $(6, 4)$, i.e. four units, then from $4p = 4$, so $p$ must be $1$. The focus will be the midpoint of the latus rectum: focus = $(4, 4)$.

Your parabola is "vertical", since the latus rectum lies on the line $y = 4$ (parallel to the x-axis). Since the point $(8, 1)$ also lies on your parabola, you should be able to tell that the parabola must open downward (in the negative $y$ direction).

With this information, and knowing $p =1$ with $p$ being the distance from the focus $(4, 4)$, you should be able to determine that the vertex $(h, k) = (5, 4)$ of your parabola (since it's one unit "up" (in positive y direction) from the focus $(4, 4))$.

Use this information to determine the equation for the parabola using the formula for a vertical parabola opening downward: $$-4p(y-k) = (x - h)^2.$$

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The corresponding equation is given by

$(x-h)^2=4p(y-k)$

where $(h,k)$ is the vertex of the parabola and $|p|$ is the distance between $(h,k)$ and the focus $F$. It can be verified that $(h,k)=(4,5)$ and $p=-1$. Plug in these two values and the equation of the parabola follows.

Note: length of latus rectum=$|4p|$.

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