The vertex of parabola is $V(3, 1)$ and the directrix is $4x + 3y = 5$.
I can't figure out the focus by using the axis of symmetry which is $3x – 4y -5 =0$.
Find the equation of the parabola given vertex and directrix
analytic geometryconic sections
Related Solutions
Hint: The axis is the line that is perpendicular to the directrix and passes through the focus. The vertex will be the midpoint of the line segment joining the focus and the intersection of the aforementioned lines.
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:
- 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.
$\qquad$
Another useful parabola feature is:
- 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.
Best Answer
Here is a straightforward way to find the unique focal point. From the directrix and the symmetry lines $$ 4x+3y=5, \>\>\>\>\>3x-4y=5$$
their intersection is $(\frac75, -\frac15)$. Since the vertex is the midpoint between the focus $(a,b)$ and the intersection point,
$$3=\frac{a+\frac75}{2},\>\>\>\>\>1=\frac{b-\frac15}{2}$$
Then, solve for the focus $(a, b)$.