analytic geometryconic sections
A comet has a parabolic orbit with the sun at the focus, when the comet is 100 million km from the sun, the line joining the sun and the comet make an angle of 60 degrees, How close will the comet get to the sun?
I could only find the horizontal distance of comet from Sun that was
$100 \cos60°\ $million km.
Parabola is symmetric wrt positive x-axis
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.
So, when you look at an body's orbit around the sun, you'll end up with an aphelion and a perihelion. These form the major axis, or the longest axis of the ellipse.
We know that $a^3/T^2 = 7.495\times10^{-6}$ from the Third Law.
$a$ represents the semi-major axis, and is half of the sum of aphelion distance (furthest) and perihelion distance (closest).
Given that $T = 27740\ \text{days}$ from 76 years, we can figure out that...
$a^3=(7.495\times10^{-6})\times(27740^2)$ --> $a \approx 17.933354$.
If aphelion $d_1$ + perihelion $d_2$ = twice the semimajor axis $2a$, and the shortest distance is $.59\ \text{a.u.}$, then we can solve for $a$:
$d_1\ \text{(far)} + d_2\ \text{(near)} = 2a\ \text{(long axis)}$
$d_1 = 2a - d_2 = 35.86 - .59 = 35.27\ \text{a.u.}$
Best Answer
Let $a$ be the minimum distance of the comet from Sun, that is the distance between focus and vertex of the parabola. The equation of the parabola, with vertex at $(0,0)$, is then: $$ x={1\over4a}y^2. $$ Substitute here the coordinates of the known position $C$ of the comet: $C=(a+50,50\sqrt3)$ (in units of $10^6\ $km) and solve for $a$.
EDIT.
A more geometrical solution (see diagram below). Let $KH$ be the directrix of the parabola so that $CH=CS=100$. Triangle $CSH$ is equilateral, hence $KS={1\over2}CH=50$ and $VS={1\over2}KS=25$.