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: Identify the axis of the parabola.
The tangent at the vertex is
\[ x-y+1=0 \]
The axis of the parabola is perpendicular to the tangent at the vertex.
Since the focus is \((0,0)\), the axis passes through \((0,0)\).

Step 2: Find the vertex.
The vertex is the foot of perpendicular from the focus \((0,0)\) to the tangent line.
For the line
\[ x-y+1=0 \]
The foot of perpendicular from \((0,0)\) is
\[ \left(-\frac12,\frac12\right) \]
So, vertex is
\[ V=\left(-\frac12,\frac12\right) \]

Step 3: Use focus-vertex-directrix relation.
The vertex is the midpoint of the focus and the foot of perpendicular from focus to the directrix.
Let that foot on the directrix be \(D\). Then,
\[ V=\frac{F+D}{2} \]
Since \(F=(0,0)\),
\[ D=2V \]
\[ D=2\left(-\frac12,\frac12\right) \]
\[ D=(-1,1) \]

Step 4: Find the directrix.
The directrix is parallel to the tangent at the vertex.
So its equation is of the form
\[ x-y+k=0 \]
Since it passes through \((-1,1)\),
\[ -1-1+k=0 \]
\[ k=2 \]
Therefore, directrix is
\[ x-y+2=0 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{x-y+2=0} \]