Question

Find the equation of the locus of all points equidistant from the point (4,2) and x-axis.
A. ${x^2} + 8x + 4y - 20 = 0$
B. ${x^2} + 8x + 4y + 20 = 0$
C. ${x^2} - 8x - 4y + 20 = 0$
D. None of these

Answer 1

Hint: First suppose the coordinate of the points. Then use the distance formula to obtain the distance between the point and (4, 2) and the x-axis. Then equate the obtained distances to obtain the required result.

Formula Used:
The distance formula of two points $(a,b),(c,d)$ is
$\sqrt {{{(c - a)}^2} + {{(d - b)}^2}} $

Complete step by step solution:
Suppose that the coordinate of the point is $P(h,k)$ and suppose (4, 2) is the point A.
It is given that $PA = k$ --(1)
Now,
$PA = \sqrt {{{(4 - h)}^2} + {{(2 - k)}^2}} $
Therefore, from equation (1) we have,
$\sqrt {{{(4 - h)}^2} + {{(2 - k)}^2}} = k$
Square both sides of the equation,
$\Rightarrow {(4 - h)^2} + {(2 - k)^2} = {k^2}$
$\Rightarrow 16 - 8h + {h^2} + 4 - 4k + {k^2} = {k^2}$
$\Rightarrow {h^2} - 8h - 4k + 20 = 0$
Therefore, the locus is ${x^2} - 8x - 4y + 20 = 0$.

Option ‘C’ is correct

Additional information:
Locus of point is the collection of all points that satisfy an equation of curve. All shapes such as circle, ellipse, hyperbola etc are defined by the locus as a set of points.

Note: Remember that the distance of a point from x axis is always the y coordinate and the distance from the y axis is always the x coordinate, no need to calculate these results. So the distance of the point $P(h,k)$ from the x-axis is h. Apply the distance formula to find distance between $P(h,k)$ and (4,2). Equate the distance to find the locus.