1. Write the hyperbola in standard form

Given $4y^2=x^2+1$ we rewrite it as $4y^2-x^2=1$ or $\frac{y^2}{(1/2)^2}-\frac{x^2}{1^2}=1.$

So this is a hyperbola of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $a^2=\frac14,\qquad b^2=1.$

2. Take a general tangent to the hyperbola

Let the tangent have equation $y=mx+c.$ Substitute into the hyperbola equation: $4(mx+c)^2=x^2+1.$

Expanding, $4m^2x^2+8mcx+4c^2=x^2+1,$ so $(4m^2-1)x^2+8mcx+(4c^2-1)=0.$

For this to represent a tangent, the quadratic in $x$ must have equal roots, so discriminant $=0$: $(8mc)^2-4(4m^2-1)(4c^2-1)=0.$

Simplify: $64m^2c^2-4(16m^2c^2-4m^2-4c^2+1)=0,$ $64m^2c^2-64m^2c^2+16m^2+16c^2-4=0,$ $16m^2+16c^2-4=0,$ $m^2+c^2=\frac14.$

Thus a tangent is $y=mx+c, \qquad m^2+c^2=\frac14.$

3. Find intercepts on the coordinate axes

For the tangent $y=mx+c,$

• $y$-intercept is at $x=0$: $A=(0,c).$

• $x$-intercept is at $y=0$: $0=mx+c \implies x=-\frac{c}{m}.$ Hence $B=\left(-\frac{c}{m},0\right).$

These are distinct points when $m\neq 0$ and $c\neq 0$.

4. Coordinates of the midpoint

Let midpoint of $AB$ be $P(x,y)$. Then $x=\frac{1}{2}\left(0-\frac{c}{m}\right)=-\frac{c}{2m},$ $y=\frac{1}{2}(c+0)=\frac{c}{2}.$

So $c=2y.$

From $x=-\frac{c}{2m},$ we get $2mx=-c=-2y,$ so $m=-\frac{y}{x}.$

5. Use the tangent condition

We know $m^2+c^2=\frac14.$ Substitute $m=-\frac{y}{x}, \qquad c=2y.$ Then $\frac{y^2}{x^2}+4y^2=\frac14.$

Multiply by $4x^2$: $4y^2+16x^2y^2=x^2.$

Rearrange: $x^2-4y^2-16x^2y^2=0.$

This is the required locus.

6. Match with the options

The locus is $x^2-4y^2-16x^2y^2=0,$ which matches Option B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.