Question

The locus of the midpoint of the chords of the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] passing through a fixed point \[\left( {\alpha ,\beta } \right)\] is a hyperbola with center is:
A) \[\left( {\dfrac{\alpha }{3},\dfrac{\beta }{3}} \right)\]
B) \[\left( {\alpha ,\beta } \right)\]
C) \[\left( {\dfrac{\alpha }{5},\dfrac{\beta }{5}} \right)\]
D) \[\left( {\dfrac{\alpha }{2},\dfrac{\beta }{2}} \right)\]

Answer 1

Hint: Here first we will assume the locus of the midpoint of the chord to be (h, k) and the write the equation of the chord with (h, k) as the midpoint and then satisfy the given points in the equation and evaluate the center of the hyperbola so formed.
The general equation of the hyperbola is:-
\[\dfrac{{{{\left( {x - {x_1}} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - {y_1}} \right)}^2}}}{{{b^2}}} = 1\]where \[\left( {{x_1},{y_1}} \right)\] is the center.
The equation of the chord with midpoint \[\left( {{x_1},{y_1}} \right)\] is given by:-
\[\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = \dfrac{{{x_1}^2}}{{{a^2}}} - \dfrac{{{y_1}^2}}{{{b^2}}}\]

Complete step-by-step answer:
The given equation of the hyperbola is:-
\[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]
Now let us assume the locus of the midpoint of the chord to be (h, k).
Now we know that the equation of the chord with midpoint \[\left( {{x_1},{y_1}} \right)\] is given by:-
\[\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = \dfrac{{{x_1}^2}}{{{a^2}}} - \dfrac{{{y_1}^2}}{{{b^2}}}\]
Hence, the equation of chord with midpoint (h, k) is given by:-
\[\dfrac{{hx}}{{{a^2}}} - \dfrac{{ky}}{{{b^2}}} = \dfrac{{{h^2}}}{{{a^2}}} - \dfrac{{{k^2}}}{{{b^2}}}\]
Now it is given that this chord passes through the point \[\left( {\alpha ,\beta } \right)\]
Hence we need to satisfy these points in the equation of the chord.
Therefore, putting these points in the above equation we get:-
\[\dfrac{{h\alpha }}{{{a^2}}} - \dfrac{{k\beta }}{{{b^2}}} = \dfrac{{{h^2}}}{{{a^2}}} - \dfrac{{{k^2}}}{{{b^2}}}\]
Solving it further we get:-
\[\dfrac{{{h^2}}}{{{a^2}}} - \dfrac{{h\alpha }}{{{a^2}}} - \dfrac{{{k^2}}}{{{b^2}}} + \dfrac{{k\beta }}{{{b^2}}} = 0\]
Taking LCM we get:-
\[\dfrac{{{h^2} - h\alpha }}{{{a^2}}} - \dfrac{{{k^2} - k\beta }}{{{b^2}}} = 0\]
Now making perfect squares we get:-
\[\dfrac{{{{\left( {h - \dfrac{\alpha }{2}} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {k - \dfrac{\beta }{2}} \right)}^2}}}{{{b^2}}} - \dfrac{{{\alpha ^2}}}{4} - \dfrac{{{\beta ^2}}}{4} = 0\]
\[\dfrac{{{{\left( {h - \dfrac{\alpha }{2}} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {k - \dfrac{\beta }{2}} \right)}^2}}}{{{b^2}}} = \dfrac{{{\alpha ^2}}}{4} + \dfrac{{{\beta ^2}}}{4}\]
Now we know that:-
The general equation of the hyperbola is:-
\[\dfrac{{{{\left( {x - {x_1}} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - {y_1}} \right)}^2}}}{{{b^2}}} = 1\]where \[\left( {{x_1},{y_1}} \right)\] is the center.
Hence the center of \[\dfrac{{{{\left( {h - \dfrac{\alpha }{2}} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {k - \dfrac{\beta }{2}} \right)}^2}}}{{{b^2}}} = \dfrac{{{\alpha ^2}}}{4} + \dfrac{{{\beta ^2}}}{4}\] is:-
\[\left( {\dfrac{\alpha }{2},\dfrac{\beta }{2}} \right)\]

Therefore, option D is correct.

Note: Students should note that the equation of the chord with given midpoint \[\left( {{x_1},{y_1}} \right)\] is given by the notation:-
\[T = {S_1}\] which on solving gives the equation as:-
\[\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = \dfrac{{{x_1}^2}}{{{a^2}}} - \dfrac{{{y_1}^2}}{{{b^2}}}\]