If the tangent drawn to the hyperbola 4 y ^ 2 = x ^ 2 + 1 intersect...

$4{y}^{2} = {x}^{2} + 1$ is $4y{y}_{1} = x{x}_{1} + 1$.

This tangent intersect coordinate axes at $A and$B$respectively, then$A\left(- \dfrac{1}{{x}_{1}}, 0\right) \text{ and } B\left(0, \dfrac{1}{4{y}_{1}}\right)$.</p><p>Let mid point is$M\left(h, k\right)$.</p><p>$2h = - \dfrac{1}{{x}_{1}} ⇒ {x}_{1} = - \dfrac{1}{2h} . . . \left(1\right)$</p><p>$2k = \dfrac{1}{{4y}_{1}} ⇒ {y}_{1} = \dfrac{1}{8k} . . . \left(2\right)$</p><p>Since, point \left({x}_{1}, {y}_{1}\right)$, lies on the hyperbola.

So, $4{{y}_{1}}^{2} = {{x}_{1}}^{2} + 1$

From equations \left(1\right)$and \left(2\right)$, we get

$4{\left(\dfrac{1}{8k}\right)}^{2} = {\left(- \dfrac{1}{2h}\right)}^{2} + 1 ⇒ \dfrac{1}{16{k}^{2}} = \dfrac{1}{4{h}^{2}} + 1$

$⇒ 4{h}^{2} = 16{k}^{2}\left(1 + 4{h}^{2}\right)$

$⇒ {x}^{2} = 4{y}^{2} + 16{x}^{2}{y}^{2}$

$⇒ {x}^{2} - 4{y}^{2} - 16{x}^{2}{y}^{2} = 0 ⇒$ Locus of $M$.