Question

The distance of the point \((1,2)\) from the directrix of the parabola \[ y^2-4x-4y+8=0 \] is:

• A. \(2\)
• B. \(1\)
• C. \(4\)
• D. \(3\)

Hint:

For the parabola \[ (y-k)^2=4a(x-h), \] remember: \[ \text{Focus}=(h+a,k), \] \[ \text{Directrix}=x=h-a. \] Once the directrix is known, finding the distance of any point from it becomes straightforward.

Answer 1

The Correct Option is B

Concept: To determine the distance of a point from the directrix of a parabola, we first convert the given parabola into its standard form. The standard form of a parabola opening towards the positive \(x\)-axis is \[ (y-k)^2=4a(x-h), \] where \[ \text{Vertex}=(h,k) \] and the directrix is \[ x=h-a. \] After obtaining the directrix, we use the perpendicular distance formula from a point to a line.

Step 1: Convert the parabola into standard form. Given \[ y^2-4x-4y+8=0. \] Grouping the \(y\)-terms, \[ y^2-4y-4x+8=0. \] Completing the square in \(y\), \[ (y^2-4y+4)-4x+8-4=0. \] \[ (y-2)^2-4x+4=0. \] \[ (y-2)^2=4(x-1). \] Comparing with \[ (y-k)^2=4a(x-h), \] we get \[ h=1,\qquad k=2,\qquad a=1. \]

Step 2: Find the equation of the directrix. The directrix of \[ (y-k)^2=4a(x-h) \] is \[ x=h-a. \] Substituting the values, \[ x=1-1=0. \] Hence the directrix is \[ x=0. \]

Step 3: Find the distance of the point from the directrix. The given point is \[ (1,2). \] The distance from \((1,2)\) to the line \[ x=0 \] is simply the horizontal distance: \[ |1-0|=1. \]

Step 4: Final Answer. Therefore, the required distance is \[ \boxed{1}. \]