Question

If A and B are fixed points of \[PA + PB = K\] , K is constant and \[K < AB\] , then find the nature of the locus of the point P.
A. Hyperbola
B. An ellipse
C. Parabola
D. A circle

Answer 1

Hint: First suppose the coordinates of A, B and P. Then obtain the values of AB, PA and PB. Then substitute the values of PA and PB in the given equation and calculate to obtain a quadratic equation. Then check the relation between \[{h^2}\] and \[ab\] to determine the nature of the locus.

Formula used:
The general form of a curve is \[a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0\]
If \[{h^2} = ab\] then the curve is parabola.
If \[{h^2} > ab\] then the curve is hyperbola.
If \[{h^2} < ab\] then the curve is ellipse.
The distance between two points \[(a,b),(c,d)\] is,
\[\sqrt {{{(c - a)}^2} + {{(d - b)}^2}} \] .

Complete step by step solution:
Suppose that the coordinate of A is (0, 0), B is (1, 0) and P is (x, y).
Then,
\[AB = \sqrt {{{\left( {1 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} \]
\[ = 1\]
It is given that \[K < AB\], we conclude that \[AB = 1\] .
So, \[K < 1.\]
Now,
\[\] \[PA = \sqrt {{{(0 - x)}^2} + {{(0 - y)}^2}} \]
\[ = \sqrt {{x^2} + {y^2}} \]
And,
\[PB = \sqrt {{{(1 - x)}^2} + {{(0 - y)}^2}} \]
\[ = \sqrt {{{\left( {1 - x} \right)}^2} + {y^2}} \]
The given equation is,
\[PA + PB = K\]
\[PA = K - PB\]
Square both sides of the equation,
\[P{A^2} = {K^2} - 2K.PB + P{B^2}\]
\[{x^2} + {y^2} = {K^2} - 2K.PB + {(1 - x)^2} + {y^2}\]
\[{x^2} + {y^2} = {K^2} - 2K.PB + 1 + {x^2} - 2x + {y^2}\]
\[2x - 1 - {K^2} = - 2K.PB\]
Square both sides of the equation,
\[4{x^2} + 1 + {K^4} - 4x + 2{K^2} - 4x{K^2} = 4{K^2}P{B^2}\]
\[4{x^2} + 1 + {K^4} - 4x + 2{K^2} - 4x{K^2} = 4{K^2}\left( {1 - 2x + {x^2} + {y^2}} \right)\]
\[4{x^2} + 1 + {K^4} - 4x + 2{K^2} - 4x{K^2} = 4{K^2} - 8x{K^2} + 4{K^2}{x^2} + 4{K^2}{y^2}\]
\[{x^2}\left( {4 - 4{K^2}} \right) - 4{K^2}{y^2} + x(4{K^2} - 4) + {K^4} - 2{K^2} + 1 = 0\]
Here,
\[a = 4 - 4{K^2},b = - 4{K^2},h = 0\]
Now,
\[ab = - 4{K^2}\left( {4 - 4{K^2}} \right)\]
\[ = 16{K^2}(1 - {K^2})\]
The term ab is negative as K<1.
Therefore,
\[{h^2} > ab\]
So, the curve is a hyperbola.

The correct option is A .

Note: To solve this type of problem students must know the relation between h, a, and b by which the nature is determined, otherwise they will be unable to solve this type of problem.