Equation of hyperbola is :

4y² = x² + 1 $$ \Rightarrow $$ $$-$$ x² + 4y² = 1

$$ \Rightarrow $$ $$\,\,\,$$ $$-$$ $${{{x^2}} \over {{1^2}}}$$ + $${{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}$$ = 1

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

Now, tangent to the curve at point (x₁, y₁) is given by

4 $$ \times $$ 2y₁$${{dy} \over {dx}}$$ = 2x₁

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

Equation of tangent at (x₁, y₁) is

y = mx + c

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

As tangent passes through (x₁, y₁)

$$ \therefore $$   y₁ = $${{{x_1}{x_1}} \over {4{y_1}}} + c$$

$$ \Rightarrow $$   C = $${{4y_1^2 - x_1^2} \over {4{y_1}}}$$ = $${1 \over {4{y_1}}}$$

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

$$ \Rightarrow $$ 4y₁y = x₁x + 1

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

Let midpoint of AB is (h, k)

$$ \therefore $$   h = $${{ - 1} \over {2{x_1}}}$$

$$ \Rightarrow $$$$\,\,\,$$ x₁ = $${{ - 1} \over {2h}}$$ & y₁ = $${1 \over {8k}}$$

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

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

$$ \Rightarrow $$ $$\,\,\,$$ 1 = $${{16{k^2}} \over {4{h^2}}}$$ + 16k²

$$ \Rightarrow $$$$\,\,\,$$ h² = 4k² + 16h² k.

So, required equation is

x² $$-$$ 4y² $$-$$ 16x² y² = 0