Given Focus S(-1, 1) and Vertex V(2, -3)
Now I have figured out two lines and those are
3x - 4y + 7 = 0
3x - 4y - 43 = 0
Now which on is the directrix.Please explain and also if there is any possible ways with graph and without graph to determine which one is the directrix.
Notice:Still haven't learnt vector and calculus.So I will not understand by those method
On
The directrix can be found by examining the distances of the line to the focus and the vertex. Note that the distance between the focus $S$ and the directrix is twice as the distance between the vertex $V$ and the directrix. Thus, we just need to check the distances between $S$ and $V$ and the two lines, respectively, using the distance formula below between the point $(p,q)$ and the line $ax+by+c=0$,
$$d=\frac{|pa+qb+c|}{\sqrt{a^2+b^2}}$$
The line $3x-4y-43=0$ is the directrix, because it has the distance 5 from the vertex $V(2,-3)$ calculated from,
$$\frac{|3\cdot 2-4(-3)-43|}{\sqrt{3^2+4^2}}=5$$
and the distance 10 from the focus $S(-1,1)$,
$$\frac{|3(-1)-4\cdot 1-43|}{\sqrt{3^2+4^2}}=10$$
i.e. twice as the distance to the vertex.
The directrix of a parabola with focus $F$ and vertex $S$ is the perpendicular to the line $(FS)$ through the point $H$ which is symmetric to $F$ w.r.t. $S$. Therefore, you first have to find the coordinates of $H$. In vector terms: $$H=F+2\overrightarrow{FV}.$$ Last, check which equation is satisfied by the coordinates of $H$.