analytic geometryconic sections
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?
Let $BE$ be the directrix
As the axis of the parabola is perpendicular to the latus rectum, the equation of the axis(OA) $y=3$
So, $c=a=3$
As the vertex O$(2,3)$ is the mid-point of $A,B, \frac{b+4}2=2\implies b=0$
So, the equation of the directrix(BE) will be $x=0$ as it is perpendicular to the axis.
If P$(h,k)$ be any point on the parabola,
the distance of $AP=\sqrt{(h-4)^2+(k-3)^2}$
the distance of P$(h,k)$ from the directrix $BE$ will be $|h|$
As the eccentricity of a parabola is $1$
So, $$\sqrt{(h-4)^2+(k-3)^2}=|h|$$
Squaring we get $(h-4)^2+(k-3)^2=h^2\implies (k-3)^2=h^2-(h-4)^2=8h-16=8(h-2)$
So, the equation of the parabola : $(y-3)^2=8(x-2)$
Let $P(7,13),F(-1,-1)$. Also, let $T$ be the intersection point of the line $y=3x-8$ with the axis of symmetry. Let $V$ be the vertex, and let $K$ be the point on the axis of symmetry such that $PK$ is perpendicular to the axis.
We use the followings (for the proof, see the end of this answer) :
(1) $PF=TF$
(2) $VT=VK$
(3) $\text{(the length of the latus rectum)}=4\times FV$
First of all, setting $T$ as $(t,3t-8)$ where $t\not= 7$ and using $(1)$ give $$(-1-7)^2+(-1-13)^2=(-1-t)^2+(-1-3t+8)^2\quad\Rightarrow\quad t=-3\quad\Rightarrow\quad T(-3,-17)$$
Hence, the axis of symmetry is the line $TF$ : $y=8x+7$. So, the line $PK$ is $y-13=(-1/8)(x-7)$, i.e. $y=-x/8+111/8$ from which $K(11/13,179/13)$ follows.
From $(2)$, since $V$ is the midpoint of the line segment $TK$, we have $V(-14/13,-21/13).$
Finally, using $(3)$, we get that the answer is $\color{red}{4\sqrt{5/13}}$.
Proof for $(1)(2)(3)$ :
We may suppose that the equation of a parabola is $y^2=4px$ where $p\gt 0$.
$\qquad\qquad\qquad$
We consider the tangent line at $A(a,b)$ where $b^2=4pa$ with $b\gt 0$. Let $B$ be the intersection point of the tangent line with $x$ axis which is the axis of symmetry. Also, let $C(p,0)$ be the focus, and let $D(a,0)$ be a point on $x$ axis such that $AD$ is perpendicular to $x$ axis. The vertex is $O(0,0)$, and let $E(p,e)$ where $e\gt 0$ be the intersection point of the parabola with the line perpendicular to $x$ axis passing through $C$.
(1)
Since the equation of the tangent line at $A$ is given by $by=2p(x+a)$, we have $B(-a,0)$, and so $$AC=\sqrt{(a-p)^2+(b-0)^2}=\sqrt{a^2-2ap+p^2+4pa}=\sqrt{(p+a)^2}=p+a=BC.$$
(2)
$OB=0-(-a)=a=OD$.
(3)
Solving $y^2=4px$ and $x=p$ gives $y=\pm 2p$, and so $e=2p$. Hence, $$\text{(the length of the latus rectum )}=2\times EC=2e=4p=4\times OC.$$
Best Answer
$h$ 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 $a=VF=4$.