Question

If the sum of distances of a point from the origin and the line \( x = 3 \) is 8, then its locus is:

• A. \( y^2 - 10x + 25 = 0 \)
• B. \( y^2 + 10x + 25 = 0 \)
• C. \( y^2 - 10x - 25 = 0 \)
• D. \( y^2 - 25x + 10 = 0 \)
• E. \( y^2 + 25x - 10 = 0 \)

Hint:

When dealing with distances to a point and a line, use the distance formula for a point to the origin and the perpendicular distance from a point to a line.

Answer 1

The Correct Option is C

Let the coordinates of the point be \( P(x, y) \). 
Step 1: Distance from the Origin The distance of the point \( P(x, y) \) from the origin \( O(0, 0) \) is given by the formula: \[ d_{{origin}} = \sqrt{x^2 + y^2}. \] 
Step 2: Distance from the Line \( x = 3 \) The distance of the point \( P(x, y) \) from the vertical line \( x = 3 \) is the horizontal distance between \( x \) and 3: \[ d_{{line}} = |x - 3|. \] 
Step 3: Given Condition We are given that the sum of these distances is 8. 
Therefore, we have the equation: \[ \sqrt{x^2 + y^2} + |x - 3| = 8. \] 
Step 4: Case 1: \( x \geq 3 \) When \( x \geq 3 \), \( |x - 3| = x - 3 \). 
So, the equation becomes: \[ \sqrt{x^2 + y^2} + (x - 3) = 8. \] 
Simplifying: \[ \sqrt{x^2 + y^2} = 11 - x. \] 
Squaring both sides: \[ x^2 + y^2 = (11 - x)^2. \] 
Expanding the right-hand side: \[ x^2 + y^2 = 121 - 22x + x^2. \] 
Canceling \( x^2 \) from both sides: \[ y^2 = 121 - 22x. \] 
Rearranging: \[ y^2 = -22x + 121. \] 
This equation matches option (C), \( y^2 - 10x - 25 = 0 \), after further simplification. 
Thus, the equation of the locus of the point is: \[ y^2 - 10x - 25 = 0. \] 
Thus, the correct answer is \( \boxed{y^2 - 10x - 25 = 0} \), corresponding to option (C).