equation of hyperbola is :

$4{y^2} = {x^2} + 1 \Rightarrow  – {x^2} + 4{y^2} = 1$

$ \Rightarrow  – \frac{{{x^2}}}{{{1^2}}} + \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = 1$

$\therefore a = 1,b = \frac{1}{2}$

Now, tangent to the curve at point $\left( {{x_1},{y_1}} \right)$ is given by.

$4 \times 2{y_1}\frac{{dy}}{{dx}} = 2{x_1}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2{x_1}}}{{8{y_1}}} = \frac{{{x_1}}}{{4{y_1}}}$

equation of tangent at $\left( {{x_1},{y_1}} \right)$ is 

$y = mx + c$

$ \Rightarrow y = \frac{{{x_1}{x_1}}}{{4{y_1}}}.x + c$

As tangent passes through ${{x_1},{y_1}}$

$\therefore {y_1} = \frac{{{x_1}{x_1}}}{{4{y_1}}} + c$

$ \Rightarrow C = \frac{{4y_1^2 – x_1^2}}{{4{y_1}}} = \frac{1}{{4{y_1}}}$

Therefore $y = \frac{{{x_1}}}{{4{y_1}}}x + \frac{1}{{4{y_1}}}$

$ \Rightarrow 4{y_1}y = {x_1}x + 1$  

which intersects at $A\left( {\frac{{ – 1}}{{{x_1}}},0} \right)$ and $y$ -axis at $B\left( {0,\frac{1}{{4{y_1}}}} \right)$

Let midpoint of $AB$ is $(h,k)$

$\therefore h = \frac{{ – 1}}{{2{x_1}}}$

$ \Rightarrow {x_1} = \frac{{ – 1}}{{2h}}\,\,\,\,\,\,{y_1} = \frac{1}{{8k}}$

Thus, $4{\left( {\frac{1}{{8k}}} \right)^2} = {\left( {\frac{{ – 1}}{{2h}}} \right)^2} + 1$

$ \Rightarrow \frac{1}{{16{k^2}}} = \frac{1}{{4{h^2}}} + 1$

$ \Rightarrow 1 = \frac{{16{k^2}}}{{4{h^2}}} + 16{k^2}\,\,\,\, \Rightarrow {h^2} = 4{k^2} + 16{h^2}k$

So, requatired equation is 

${x^2} – 4{y^2}16{x^2}{y^2} = 0$