Question

The locus of the mid-points of the chord passing through a fixed point ($\alpha $,$\beta $) of the hyperbola $\dfrac{{x_{}^2}}{{a_{}^2}} - \dfrac{{y_{}^2}}{{b_{}^2}} = 1$ is a hyperbola whose centre is
a) $\left( {\dfrac{\alpha }{3},\dfrac{\beta }{3}} \right)$
b) $(\alpha ,\beta )$
c) $\left( {\dfrac{\alpha }{5},\dfrac{\beta }{5}} \right)$
d) $\left( {\dfrac{\alpha }{2},\dfrac{\beta }{2}} \right)$

Answer 1

Hint: First we assume the locus of the middle point of the hyperbola and we use the formula for the equation of the hyperbola.
Then we do some simplification and we get the answer.

Formula used: Equation of chord of hyperbola, $\dfrac{{x{x_1}}}{{a_{}^2}} - \dfrac{{y{y_1}}}{{b_{}^2}} = \dfrac{{x_1^2}}{{a_{}^2}} - \dfrac{{y_{}^2}}{{b_{}^2}}$

Complete step-by-step answer:
Let us assume that the locus of the middle point of the chord of the hyperbola is \[\left( {h,k} \right)\]
By applying the formula of equation of chord of hyperbola and putting the values of ${x_1} = h$ and ${y_1} = k$ we get-
$\dfrac{{xh}}{{a_{}^2}} - \dfrac{{yk}}{{b_{}^2}} = \dfrac{{h_{}^2}}{{a_{}^2}} - \dfrac{{k_{}^2}}{{b_{}^2}}$
Moving the terms to the left hand side for making it zero we get,
$\dfrac{{xh}}{{a_{}^2}} - \dfrac{{yk}}{{b_{}^2}} - \dfrac{{h_{}^2}}{{a_{}^2}} + \dfrac{{k_{}^2}}{{b_{}^2}} = 0$
On multiplying ($ - $) sign, we can write it as
$\dfrac{{h_{}^2}}{{a_{}^2}} - \dfrac{{xh}}{{a_{}^2}} - \dfrac{{k_{}^2}}{{b_{}^2}} + \dfrac{{yk}}{{b_{}^2}} = 0$
Now taking common terms of ${a^2}$ and ${b^2}$ we get,
$\dfrac{1}{{a_{}^2}}(h_{}^2 - hx) - \dfrac{1}{{b_{}^2}}(k_{}^2 - ky) = 0....\left( 1 \right)$
Now we will make the equation in whole square format
We have to add and sub $\dfrac{{x_{}^2}}{{4a_{}^2}}$ and $\dfrac{{y_{}^2}}{{4b_{}^2}}$ we get,
$\dfrac{1}{{a_{}^2}}\left[ {\left( {h_{}^2 - 2.h.\dfrac{x}{2} + \dfrac{{x_{}^2}}{4}} \right) - \dfrac{{x_{}^2}}{{4a_{}^2}}} \right] - \dfrac{1}{{b_{}^2}}\left[ {\left( {k_{}^2 - 2.k.\dfrac{y}{2} + \dfrac{{y_{}^2}}{4}} \right) + \dfrac{{y_{}^2}}{{4b_{}^2}}} \right] = 0$
So the above equation turns into the formula of $(a - b)_{}^2 = a_{}^2 - 2ab + b_{}^2$
$\dfrac{1}{{a_{}^2}}\left( {h - \dfrac{x}{2}} \right)_{}^2 - \dfrac{{x_{}^2}}{{4a_{}^2}} - \dfrac{1}{{b_{}^2}}\left( {k_{}^{} - \dfrac{y}{2}} \right)_{}^2 + \dfrac{{y_{}^2}}{{4b_{}^2}} = 0$
We can write \[x\] and \[y\] term as in RHS,
$\dfrac{1}{{a_{}^2}}\left( {h - \dfrac{x}{2}} \right)_{}^2 - \dfrac{1}{{b_{}^2}}\left( {k - \dfrac{y}{2}} \right)_{}^2 = \dfrac{{x_{}^2}}{{4a_{}^2}} - \dfrac{{y_{}^2}}{{4b_{}^2}}$
Taking $\dfrac{1}{4}$ as common from the right side we get-
$\dfrac{1}{{a_{}^2}}\left( {h - \dfrac{x}{2}} \right)_{}^2 - \dfrac{1}{{b_{}^2}}\left( {k_{}^{} - \dfrac{y}{2}} \right)_{}^2 = \dfrac{1}{4}\left( {\dfrac{{x_{}^2}}{{a_{}^2}} - \dfrac{{y_{}^2}}{{b_{}^2}}} \right)....\left( 2 \right)$
Now we have to put the values of $\left( {\alpha ,\beta } \right)$ in place of $\left( {h,k} \right)$ in the given equation \[\left( 1 \right)\]
Since the chord passes through this fixed point of the hyperbola so we can write
$\dfrac{1}{{a_{}^2}}(\alpha _{}^2 - \alpha x) - \dfrac{1}{{b_{}^2}}(\beta _{}^2 - \beta y) = 0$
From here we get that $x = \alpha $and $y = \beta $
Putting this value in equation \[\left( 2 \right)\] we get
\[\dfrac{1}{{a_{}^2}}\left( {h - \dfrac{\alpha }{2}} \right)_{}^2 - \dfrac{1}{{b_{}^2}}\left( {k_{}^{} - \dfrac{\beta }{2}} \right)_{}^2 = \dfrac{1}{4}\left( {\dfrac{{\alpha _{}^2}}{{a_{}^2}} - \dfrac{{\beta _{}^2}}{{b_{}^2}}} \right)\]
So this is an equation of hyperbola and now converting $\left( {h,k} \right)$ into $\left( {x,y} \right)$ we get-
\[\dfrac{1}{{a_{}^2}}\left( {x - \dfrac{\alpha }{2}} \right)_{}^2 - \dfrac{1}{{b_{}^2}}\left( {y - \dfrac{\beta }{2}} \right)_{}^2 = \dfrac{1}{4}\left( {\dfrac{{\alpha _{}^2}}{{a_{}^2}} - \dfrac{{\beta _{}^2}}{{b_{}^2}}} \right)\]
Here we can write,
Centre of coordinates of the hyperbola is $\left( {\dfrac{\alpha }{2},\dfrac{\beta }{2}} \right)$

Therefore the correct option is d.

Note: The centre of hyperbola gets intersected by two lines and the tangent to the centre is called asymptotes of the hyperbola.
The value of $\dfrac{{x_{}^2}}{{a_{}^2}} - \dfrac{{y_{}^2}}{{b_{}^2}} = 1$ can be positive, negative or zero, and it depend on the point where it lies that is within, on or outside of the hyperbola.