Question

A point moves in such a way that the difference of its distance from two point \[\left( {8,0} \right)\] and \[\left( { - 8,0} \right)\] always remains \[4\]. Then the locus of the point is
A. A circle
B. A parabola
C. An ellipse
D. A hyperbola

Answer 1

Hint: Let at an instant, the coordinates of the moving point \[P\] be \[\left( {h,k} \right)\] and the coordinates of the two fixed points \[A\] and \[B\] are \[\left( {8,0} \right)\] and \[\left( { - 8,0} \right)\]. Find the length of the line segments \[PA\] and \[PB\] and make an equation using the given condition and simplify the equation. After simplification replace \[h\] by \[x\] and \[k\] by \[y\]. Then identify the conic.

Formula Used:
Distance between two points \[A\left( {{x_1},{y_1}} \right)\] and \[B\left( {{x_2},{y_2}} \right)\] is \[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Complete step-by-step solution:
Let the coordinates of the variable point \[P\] be \[\left( {h,k} \right)\] and the coordinates of the two fixed points \[A\] and \[B\] are given as \[\left( {8,0} \right)\] and \[\left( { - 8,0} \right)\].
Let us find the lengths of the line segments \[PA\] and \[PB\]
For \[PA\], \[{x_1} = h,{y_1} = k,{x_2} = 8,{y_2} = 0\]
So, \[PA = \sqrt {{{\left( {h - 8} \right)}^2} + {{\left( {k - 0} \right)}^2}} = \sqrt {{h^2} - 16h + 64 + {k^2}} \]
For \[PB\], \[{x_1} = h,{y_1} = k,{x_2} = - 8,{y_2} = 0\]
So, \[PB = \sqrt {{{\left( {h + 8} \right)}^2} + {{\left( {k - 0} \right)}^2}} = \sqrt {{h^2} + 16h + 64 + {k^2}} \]
It is said in the question that the difference of the moving point from the two fixed points is \[4\].
So, we have
\[\left| {PA - PB} \right| = 4\]
\[ \Rightarrow PA - PB = \pm 4\]
Let us assume that \[PA - PB = 4\]
Substituting the values of \[PA\] and \[PB\] in the equation, we get
\[ \Rightarrow \sqrt {{h^2} - 16h + 64 + {k^2}} - \sqrt {{h^2} + 16h + 64 + {k^2}} = 4\]
\[ \Rightarrow \sqrt {{h^2} - 16h + 64 + {k^2}} = 4 + \sqrt {{h^2} + 16h + 64 + {k^2}} \]
Squaring both sides, we get
\[ \Rightarrow {\left( {\sqrt {{h^2} - 16h + 64 + {k^2}} } \right)^2} = {\left( {4 + \sqrt {{h^2} + 16h + 64 + {k^2}} } \right)^2}\]
Use the identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[ \Rightarrow {h^2} - 16h + 64 + {k^2} = 16 + 8\sqrt {{h^2} + 16h + 64 + {k^2}} + {h^2} + 16h + 64 + {k^2}\]
Cancel the terms \[{h^2},{k^2}\] and \[64\] from both sides.
\[ \Rightarrow - 16h = 16 + 8\sqrt {{h^2} + 16h + 64 + {k^2}} + 16h\]
\[ \Rightarrow - 32h - 16 = 8\sqrt {{h^2} + 16h + 64 + {k^2}} \]
Take out \[\left( { - 16} \right)\] as common on left side.
\[ \Rightarrow - 16\left( {2h + 1} \right) = 8\sqrt {{h^2} + 16h + 64 + {k^2}} \]
Cancel \[8\] from both sides.
\[ \Rightarrow 2\left( {2h + 1} \right) = - \sqrt {{h^2} + 16h + 64 + {k^2}} \]
Square both sides.
\[ \Rightarrow {\left\{ {2\left( {2h + 1} \right)} \right\}^2} = {\left\{ { - \sqrt {{h^2} + 16h + 64 + {k^2}} } \right\}^2}\]
\[ \Rightarrow 4{\left( {2h + 1} \right)^2} = {h^2} + 16h + 64 + {k^2}\]
Use the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[ \Rightarrow 4\left( {4{h^2} + 4h + 1} \right) = {h^2} + 16h + 64 + {k^2}\]
\[ \Rightarrow 16{h^2} + 16h + 4 = {h^2} + 16h + 64 + {k^2}\]
\[ \Rightarrow 16{h^2} - {h^2} - {k^2} + 16h - 16h + 4 - 64 = 0\]
\[ \Rightarrow 15{h^2} - {k^2} - 60 = 0\]
Replacing \[h\] by \[x\] and \[k\] by \[y\], we get
\[15{x^2} - {y^2} - 60 = 0\]
This is the equation of the locus of \[P\left( {h,k} \right)\].
Comparing the equation of the locus with the general form of a conic \[A{x^2} + Bxy + C{y^2} + Dx + Ey + F = 0\], we get
\[A = 15,B = 0,C = - 1,D = 0,E = 0,F = - 60\]
Now, \[{B^2} - 4AC = {\left( 0 \right)^2} - 4\left( {15} \right)\left( { - 1} \right) = 0 + 60 = 60 > 0\]
So, \[{B^2} - 4AC > 0\]
Thus, the equation represents a hyperbola.
Hence, option D is correct.

Note: Every second degree equation in two variables represents a conic. Many students can’t remember the conditions for the equations to be of a parabola, an ellipse, a hyperbola and a circle. So, they can’t identify the equation of the conic. For parabola, \[{B^2} - 4AC = 0\] and \[A = 0\] or \[C = 0\]. For ellipse, \[{B^2} - 4AC < 0\] and \[A \ne C\]. For hyperbola, \[{B^2} - 4AC > 0\], For circle, \[{B^2} - 4AC < 0\] and \[A = C\].