Question

The coordinates of the points A and B are $(a,0)$ and $( - a,0)$ respectively. If a point P moves so that $P{A^2} - P{B^2} = 2{k^2}$ , when k is constant, then find the equation to the locus of the point P.
A. $2ax + {k^2} = 0$
B. $2ax - {k^2} = 0$
C. $ax + 2{k^2} = 0$
D. $ax - 2{k^2} = 0$

Answer 1

Hint: First consider the coordinate of the point P. Then use the distance formula to obtain the distance between the point P and A, and that point P and B. Then substitute the values of PA and PB in the given equation to obtain the required locus.

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 $P(x,y)$ .
It is given that $P{A^2} - P{B^2} = 2{k^2}$--(1)
Now, $PA = \sqrt {{{(a - x)}^2} + {{(0 - y)}^2}} $
$P{A^2} = {(a - x)^2} + {y^2}$
And, $PB = \sqrt {{{( - a - x)}^2} + {{(0 - y)}^2}} $
$P{B^2} = {(a + x)^2} + {y^2}$
Now, from the equation (1) we have,
${(a - x)^2} + {y^2} - \left[ {{{(a + x)}^2} + {y^2}} \right] = 2{k^2}$
$\Rightarrow {a^2} - 2ax + {x^2} + {y^2} - \left[ {{a^2} + 2ax + {x^2} + {y^2}} \right] = 2{k^2}$
$\Rightarrow - 4ax = 2{k^2}$
$\Rightarrow 2ax + {k^2} = 0$

Option ‘A’ is correct

Note: Always consider the coordinates of P other than (h, k) to avoid confusion between the constant k and the coordinate k. Here using the distance formula of two points $(a,b),(c,d)$ it becomes easy to equate the given coordinates. If the diameter line is also bisecting the sides and is perpendicular then we have to find all equations of sides and equations of circle and then the coordinates of vertices.