Question

If \[A\] and \[B\] are two fixed points and \[P\] is a variable point such that \[PA + PB = 4\], where \[AB < 4\], what is the locus of\[P\]?
A. A parabola
B. An ellipse
C. A hyperbola
D. None of these

Answer 1

Hint: Let the coordinates of the variable point \[P\] be \[\left( {h,k} \right)\] and the coordinates of the two fixed points \[A\] and \[B\] be \[\left( {a,b} \right)\] and \[\left( {c,d} \right)\]. Find the length of the line segments \[PA\] and \[PB\] and substitute these in the given equation \[PA + PB = 4\] and \[AB < 4\].

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 \[B\] be \[\left( {a,b} \right)\] and \[\left( {c,d} \right)\].
Let us find the lengths of the line segments \[PA\] and \[PB\]
For \[PA\], \[{x_1} = h,{y_1} = k,{x_2} = a,{y_2} = b\]
So, \[PA = \sqrt {{{\left( {a - h} \right)}^2} + {{\left( {b - k} \right)}^2}} \]
For \[PB\], \[{x_1} = h,{y_1} = k,{x_2} = c,{y_2} = d\]
So, \[PB = \sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \]
Given that \[PA + PB = 4\]
Substituting the values of \[PA\] and \[PB\] in the equation, we get
\[\sqrt {{{\left( {a - h} \right)}^2} + {{\left( {b - k} \right)}^2}} + \sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} = 4\]
\[ \Rightarrow \sqrt {{{\left( {a - h} \right)}^2} + {{\left( {b - k} \right)}^2}} = 4 - \sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \]
Squaring both sides, we get
\[ \Rightarrow {\left( {\sqrt {{{\left( {a - h} \right)}^2} + {{\left( {b - k} \right)}^2}} } \right)^2} = {\left( {4 - \sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} } \right)^2}\]
Use the identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {\left( {a - h} \right)^2} + {\left( {b - k} \right)^2} = 16 - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} + {\left( {c - h} \right)^2} + {\left( {d - k} \right)^2}\]
Again, use the identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {a^2} - 2ah + {h^2} + {b^2} - 2bk + {k^2} = 16 - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} + {c^2} - 2ch + {h^2} + {d^2} - 2dk + {k^2}\]
Cancel the terms \[{h^2}\] and \[{k^2}\] from both sides.
\[ \Rightarrow {a^2} - 2ah + {b^2} - 2bk = 16 - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} + {c^2} - 2ch + {d^2} - 2dk\]
\[ \Rightarrow {a^2} + {b^2} - {c^2} - {d^2} - 2ah - 2bk + 2ch + 2dk - 16 = - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \]
\[ \Rightarrow {a^2} + {b^2} - {c^2} - {d^2} + 2\left( {c - a} \right)h + 2\left( {d - b} \right)k - 16 = - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \]
The points \[A\left( {a,b} \right)\] and \[B\left( {c,d} \right)\] are fixed. It means \[a,b,c,d\] are constants.
Let \[{a^2} + {b^2} - {c^2} - {d^2} - 16 = m\], \[2\left( {c - a} \right) = p\] and \[2\left( {d - b} \right) = q\]
Then the equation becomes
\[m + ph + qk = - 8\sqrt {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \]
Squaring both sides, we get
\[ \Rightarrow {\left( {m + ph + qk} \right)^2} = 64\left\{ {{{\left( {c - h} \right)}^2} + {{\left( {d - k} \right)}^2}} \right\}\]
Use the formula \[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\] and \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {m^2} + {p^2}{h^2} + {q^2}{k^2} + 2mph + 2pqhk + 2mqk = 64\left( {{c^2} - 2ch + {h^2} + {d^2} - 2dk + {k^2}} \right)\]
\[ \Rightarrow {m^2} + {p^2}{h^2} + {q^2}{k^2} + 2mph + 2pqhk + 2mqk = 64{c^2} - 128ch + 64{h^2} + 64{d^2} - 128dk + 64{k^2}\]
Arrange the terms.
\[ \Rightarrow {p^2}{h^2} - 64{h^2} + 2pqhk + {q^2}{k^2} - 64{k^2} + 2mph + 128ch + 2mqk + 128dk + {m^2} - 64{c^2} - 64{d^2} = 0\]
\[ \Rightarrow \left( {{p^2} - 64} \right){h^2} + 2pqhk + \left( {{q^2} - 64} \right){k^2} + 2\left( {mp + 64c} \right)h + 2\left( {mq + 64d} \right)k + \left( {{m^2} - 64{c^2} - 64{d^2}} \right) = 0\]
Replacing \[h\] by \[x\] and \[k\] by \[y\], we get
\[ \Rightarrow \left( {{p^2} - 64} \right){x^2} + 2pqxy + \left( {{q^2} - 64} \right){y^2} + 2\left( {mp + 64c} \right)x + 2\left( {mq + 64d} \right)y + \left( {{m^2} - 64{c^2} - 64{d^2}} \right) = 0\]
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 \[A = {p^2} - 64\], \[B = 2pq\], \[C = {q^2} - 64\], \[D = 2\left( {mp + 64c} \right)\], \[E = 2\left( {mq + 64d} \right)\], \[F = {m^2} - 64{c^2} - 64{d^2}\]
Now, \[{B^2} - 4AC = {\left( {2pq} \right)^2} - 4\left( {{p^2} - 64} \right)\left( {{q^2} - 64} \right)\]
\[ = 4{p^2}{q^2} - 4{p^2}{q^2} + 256{p^2} + 256{q^2} - 16384\]
\[ = 256{p^2} + 256{q^2} - 16384\]
\[ = 256\left( {{p^2} + {q^2} - 64} \right)\]
We assumed that \[2\left( {c - a} \right) = p\] and \[2\left( {d - b} \right) = q\]
\[ = 256\left[ {{{\left\{ {2\left( {c - a} \right)} \right\}}^2} + {{\left\{ {2\left( {d - b} \right)} \right\}}^2} - 64} \right]\]
\[ = 256\left\{ {4{{\left( {c - a} \right)}^2} + 4{{\left( {d - b} \right)}^2} - 64} \right\}\]
\[ = 256\left[ {4\left\{ {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2} - 16} \right\}} \right]\]
\[ = 1024\left\{ {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2} - 16} \right\}\]
Now, it is given that \[AB < 4\]
\[ \Rightarrow \sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}} < 4\]
\[ \Rightarrow {\left( {a - c} \right)^2} + {\left( {b - d} \right)^2} < 16\]
\[ \Rightarrow {\left( {a - c} \right)^2} + {\left( {b - d} \right)^2} - 16 < 0\]
So, \[{B^2} - 4AC < 0\] and \[A \ne C\], in general.
Thus, the equation represents an ellipse.
Hence, option B 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 \[A = C\].