Question

The focal distance of a point (5,5) on the parabola \( x^2-2x-4y+5=0 \) is

• A. 5
• B. 8
• C. 10
• D. 12

Hint:

For \( (x-h)^2 = 4a(y-k) \), focal distance = \( |y_{\text{point}} - y_{\text{directrix}}| \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Convert the equation to standard form to identify the axis and directrix. The focal distance of a point is its distance from the focus, which equals its perpendicular distance from the directrix (definition of parabola).
Step 2: Detailed Explanation:
Equation: \( x^2 - 2x = 4y - 5 \).
Complete the square for x:
\( (x-1)^2 - 1 = 4y - 5 \)
\( (x-1)^2 = 4y - 4 \)
\( (x-1)^2 = 4(1)(y-1) \).
Standard form \( X^2 = 4aY \) where \( a=1 \), vertex at \( (1,1) \).
This is an upward-opening parabola.
Directrix equation in standard form is \( Y = -a \).
\( y - 1 = -1 \implies y = 0 \).
The directrix is the x-axis (\( y=0 \)).
Focal distance of a point P on the parabola = Distance of P from directrix. Point P is \( (5, 5) \).
Distance from \( y=0 \) is simply the y-coordinate, which is 5.
Alternatively, Focus S is \( (1, 1+1) = (1, 2) \).
Distance \( SP = \sqrt{(5-1)^2 + (5-2)^2} = \sqrt{16+9} = 5 \).

Step 3: Final Answer:
The focal distance is 5.