Question

Suppose a parabola with focus at \((0,0)\) has
\[ x-y+1=0 \] as its tangent at the vertex. Then the equation of its directrix is

• A. \(x-y+2=0\)
• B. \(x-y-2=0\)
• C. \(x-y+3=0\)
• D. \(x-y+4=0\)

Hint:

For a parabola, the tangent at the vertex is perpendicular to the axis. The vertex is the midpoint between the focus and the corresponding point on the directrix.

Answer 1

The Correct Option is A

Step 1: Use the vertex-tangent property.
The tangent at the vertex of a parabola is perpendicular to the axis, and the vertex is the foot of the perpendicular from the focus to this tangent.
Step 2: Drop a perpendicular from the focus.
The focus is $(0,0)$ and the tangent is $x-y+1=0$. The foot of the perpendicular from $(0,0)$ to this line is the vertex $V=\left(-\dfrac12,\dfrac12\right)$.
Step 3: Use the midpoint relation for the directrix.
The vertex is the midpoint between the focus and the foot $D$ of the focus on the directrix, so $D=2V-F$.
Step 4: Compute $D$.
$D=2\left(-\dfrac12,\dfrac12\right)-(0,0)=(-1,1)$.
Step 5: The directrix is parallel to the tangent.
So it has the form $x-y+k=0$. Passing through $(-1,1)$: $-1-1+k=0$, so $k=2$.
Step 6: Write the directrix.
The directrix is $x-y+2=0$.
\[ \boxed{x-y+2=0} \]