Find Vertex when Focus and point on directrix of Parabola is given.

Question

Focus is $(2,3)$, point on directrix is $(-3,2)$. Parabola touches $x$-axis. Find vertex.

I would be thankful if someone could help me with this problem.

Answer 1

HINT:

The equation of the directrix can be written as $$\dfrac{y-2}{x+3}=m$$

As the eccentricity is $1,$

the distance of any point $P(h,k)$ on the parabola from the focus = the distance from the directrix.

Now as $y=0$ is a tangent of the parabola, put $y=0$ in the relation derived above to form a Quadratic Equation in $x$ whose each root represents the abscissa of intersection.

Now for tangency, both root should be same.

Answer 2

After some inspection and algebraic manipulation we find that the directrix is $x+y+1=0$, and the axis of symmetry is $x-y+1=0$, both meeting conveniently at $(-1,0)$. The midpoint between this and the focus $F(2,3)$ is the vertex $\color{red}{V(0.5,1.5)}$. (For a more rigorous analysis, see section titled "In Greater Detail" below.)

The required parabola is thus tilted at $45^\circ$ and can be described in several forms as shown below:

$$(x-2)^2+(y-3)^2=\frac {(x+y+1)^2}2$$ or $$(x-y)^2-10x-14y+25=0$$ or $$(x-y-5)^2=24y$$ The $x-$axis ($y=0$) is tangential to the parabola at $(5,0)$.

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Note that $x=-1$ is also tangential to the parabola (at $(-1,6)$), and, together with the $y=0$ (the $x-$axis) form a set of perpendicular tangents, which intersect at a point on the directrix, this being a property of the parabola. In this case the point of intersection is $(-1,0)$, which as was ascertained above, lies on the directrix.

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In Greater Detail

Let focus be $F=(2,3)$ and point on directrix be $D=(-3,2)$ (values as given).

Let the point $R$ be the foot of the perpendicular from $F$ to the directrix. As the directrix is perpendicular to the axis of symmetry, $\angle DRF=90^\circ$. It follows that the locus of $R$ is a circle with diameter $DF$, i.e. $(x+\frac 12)^2+(y-\frac 32)^2=\frac {13}2$. At $y=0$, $x=0,-1$, ie. the circle crosses the $x-$axis at $(-1,0), (0,0)$. $DR$ is part of the directrix.

It is given that the $x-$axis is a tangent to the required parabola. We also know that for any parabola, two perpendicular tangents cross at a point on the directrix. Hence the perpendicular tangent to the $x-$axis must be a line parallel to the $y-$axis, i.e. the intersection point (which lies on the direcrix) must also lie on the $x-$axis.

Given the above we conclude that $R$ lies on the $x-$axis, and can be either $R_1(-1,0)$ or $R_2(0,0)$.

Let point $P(x,y)$ be a general point on the required parabola. By the basic definition of a parabola, $FP=PG$ where $G$ is the foot of the perpendicular from $P$ to the directrix. $FP^2=PG^2$ gives the equation of the parabola.

• If $R=R_1$, the parabola is $$(x-2)^2+(y-3)^2=\frac {(x+y-1)^2}2$$ and at $y=0$, we have $(x-5)^2=0$, i.e. the $x-$axis is a tangent to the parabola, with $(5,0)$ as the tangential point.

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• If $R=R_2$, the parabola is $$(x-2)^2+(y-3)^2=\frac {(2x+3y)^2}{13}$$ and at $y=0$, we have $(x-2)^2+9=\frac {4x^2}{13} $ which reduces to $9x^2-52x+169=0$, a quadratic with no real roots, i.e. the parabola does not touch or cross the $x-$axis.

Hence we conclude that $R=R_1(-1,0)$ and the required parabola is $$(x-2)^2+(y-3)^2=\frac {(x+y-1)^2}2$$ with the $x-$axis tangential to it at $(5,0)$. The vertex of the parabola is the midpoint of $RF$, i.e. $\color{red}{V\big(\frac 12, \frac 32\big)}$.

Answer 3

Here is a geometric way; I leave it to you to translate it to algebra.

You are given the focus $F$, a point $D$ on the directrix and a tangent $a$. For a parabola we have:

1. Mirroring the focus across a tangent yields a point on the directrix.

So mirroring $F$ across $a$ results in a point $F'$, and the line $DF'$ is the directrix.

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Another useful parabola feature is:

2. The parabola vertex lies halfway between the focus and the closest point of the directrix.

Therefore consider a straight line through $F$ that intersects the directrix orthogonally in a point $P$. The midpoint between $P$ and $F$ is the sought vertex $V$.

With the given coordinates, $P$ happens to land on the tangent $a$, but that is mere coincidence.