If the tangent drawn to the hyperbola $4y^2=x^2+1$ intersect the co-ordinate axes at the distinct points $A$and $B$, then the locus of the mid point of $AB$ is 

We have, $4y^2=x^2+1$                      …(i)

The tangent to (i) at $\left(x_1,y_1\right)$ is given by $4y{y}_1=x{x}_1+1$

According to question, $A$ $\equiv \left(\frac{-1}{x_1},0\right),B\equiv \left(0,\frac{1}{4y_1}\right)$

Let mid point of $AB$ be $M$$(h, k).$

Then, $\frac{-1}{x_1}=2h\Rightarrow x_1=\frac{-1}{2h}$ and $\frac{1}{4y_1}=2k\Rightarrow y_1=\frac{1}{8k}$

$\because \left(x_1,y_1\right)$lies on (i)

$\begin{array}{r}\therefore 4{\left(\frac{1}{8k}\right)}^2={\left(\frac{-1}{2h}\right)}^2+1\Rightarrow \frac{4\mid \times 1}{64k^2}=\frac{1}{4h^2}+1\\ \Rightarrow \frac{1}{16k^2}=\frac{1}{4h^2}+1\Rightarrow h^2=4k^2+16h^2{k}^2\end{array}$

  $\therefore$      Locus of mid point of $AB$ is

$x^2=4y^2+16x^2{y}^2$ or $x^2-4y^2-16x^2{y}^2=0$